interviews | Everything you need to know to get the job | Learning library

 by   kdn251 Java Version: Current License: MIT

kandi X-RAY | interviews Summary

interviews is a Java library typically used in Tutorial, Learning, Example Codes, LeetCode applications. interviews has no bugs, it has no vulnerabilities, it has a Permissive License and it has medium support. However interviews build file is not available. You can download it from GitHub.
Your personal guide to Software Engineering technical interviews. Video solutions to the following interview problems with detailed explanations can be found here. Maintainer - Kevin Naughton Jr.
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                      kandi-support Support

                        summary
                        interviews has a medium active ecosystem.
                        summary
                        It has 59378 star(s) with 12514 fork(s). There are 2621 watchers for this library.
                        summary
                        It had no major release in the last 6 months.
                        summary
                        There are 31 open issues and 20 have been closed. On average issues are closed in 35 days. There are 78 open pull requests and 0 closed requests.
                        summary
                        It has a neutral sentiment in the developer community.
                        summary
                        The latest version of interviews is current.
                        interviews Support
                          Best in #Learning
                            Average in #Learning
                            interviews Support
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                                  kandi-Quality Quality

                                    summary
                                    interviews has 0 bugs and 0 code smells.
                                    interviews Quality
                                      Best in #Learning
                                        Average in #Learning
                                        interviews Quality
                                          Best in #Learning
                                            Average in #Learning

                                              kandi-Security Security

                                                summary
                                                interviews has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.
                                                summary
                                                interviews code analysis shows 0 unresolved vulnerabilities.
                                                summary
                                                There are 0 security hotspots that need review.
                                                interviews Security
                                                  Best in #Learning
                                                    Average in #Learning
                                                    interviews Security
                                                      Best in #Learning
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                                                          kandi-License License

                                                            summary
                                                            interviews is licensed under the MIT License. This license is Permissive.
                                                            summary
                                                            Permissive licenses have the least restrictions, and you can use them in most projects.
                                                            interviews License
                                                              Best in #Learning
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                                                                interviews License
                                                                  Best in #Learning
                                                                    Average in #Learning

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                                                                        summary
                                                                        interviews releases are not available. You will need to build from source code and install.
                                                                        summary
                                                                        interviews has no build file. You will be need to create the build yourself to build the component from source.
                                                                        summary
                                                                        Installation instructions are not available. Examples and code snippets are available.
                                                                        summary
                                                                        interviews saves you 5461 person hours of effort in developing the same functionality from scratch.
                                                                        summary
                                                                        It has 11447 lines of code, 777 functions and 516 files.
                                                                        summary
                                                                        It has medium code complexity. Code complexity directly impacts maintainability of the code.
                                                                        interviews Reuse
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                                                                                  Top functions reviewed by kandi - BETA
                                                                                  kandi has reviewed interviews and discovered the below as its top functions. This is intended to give you an instant insight into interviews implemented functionality, and help decide if they suit your requirements.
                                                                                  • Decodes a string .
                                                                                    • Returns the minimal window .
                                                                                      • Get the vertical order of the node in the tree
                                                                                        • Generate matrix .
                                                                                          • Returns a list of integers .
                                                                                            • Returns the level order in the tree rooted at the tree
                                                                                              • Merge k lists .
                                                                                                • Group an array of Strings
                                                                                                  • Determine how many substrings are unique
                                                                                                    • Return the minimum cost of an array of costs .
                                                                                                      Get all kandi verified functions for this library.
                                                                                                      Get all kandi verified functions for this library.

                                                                                                      interviews Key Features

                                                                                                      Everything you need to know to get the job.

                                                                                                      interviews Examples and Code Snippets

                                                                                                      default
                                                                                                      Javascriptdot imgLines of Code : 40dot imgno licencesLicense : No License
                                                                                                      copy iconCopy
                                                                                                      
                                                                                                                                          Now why would we care?
                                                                                                      
                                                                                                      expenses: Expense[]; ngOnInit() { this.getExpenses() .subscribe(expenses => { this.expenses = expenses; }); }
                                                                                                      
                                                                                                      expenses: Expense[] = []; filter = "food"; ngOnInit() { this.getExpenses() .subscribe(expenses => { this.expenses = expenses.filter(e => e.type === this.filter); }); this.getFilter() .subscribe(filter => { this.filter = filter; this.expenses = this.expenses.filter(e => e.type === filter); }); }
                                                                                                      
                                                                                                      expenses: Expense[] = []; ngOnInit() { this.getExpenses() .combineLatest(this.getFilter()) .subscribe(([expenses, filter]) => { this.expenses = expenses.filter(e => e.type === filter); }); }
                                                                                                      
                                                                                                      expenses$: Observable; ngOnInit() { this.expenses$ = this.getExpenses() .combineLatest(this.getFilter()) .map(([expenses, filter]) => expenses.filter(e => e.type === filter)); }
                                                                                                      
                                                                                                      default
                                                                                                      Javascriptdot imgLines of Code : 30dot imgno licencesLicense : No License
                                                                                                      copy iconCopy
                                                                                                      
                                                                                                                                          import { switchMap } from 'rxjs/operators'; import { Component, OnInit } from '@angular/core'; import { Router, ActivatedRoute, ParamMap } from '@angular/router'; import { Observable } from 'rxjs'; import { EmployeeService } from '../employee.service'; import { Employee} from '../employee; @Component({ selector: 'app-employee-detail', templateUrl: './employee-detail.component.html', styleUrls: ['./employee-detail.component.css'] }) export class EmployeeDetailComponent implements OnInit { employee$: Observable; constructor( private route: ActivatedRoute, private router: Router, private service: EmployeeService ) {} ngOnInit() { this.employee$ = this.route.paramMap.pipe( switchMap((params: ParamMap) => this.service.getEmployee(params.get('id'))) ); } } }
                                                                                                      
                                                                                                      copy iconCopy
                                                                                                      
                                                                                                                                          @Component({ selector: "async-pipe", template: `  AsyncPipe 
                                                                                                      

                                                                                                      {{ observable | async }}

                                                                                                      {{ observable | async }}

                                                                                                      (1) ` }) class AsyncPipeComponent { observable: Observable; constructor() { this.observable = this.getObservable(); } getObservable() { return Observable.interval(1000) .take(10) .map(v => v * v); } }
                                                                                                      Function used to perform the operation .
                                                                                                      javascriptdot imgLines of Code : 21dot imgno licencesLicense : No License
                                                                                                      copy iconCopy
                                                                                                      
                                                                                                                                          function performOperation(secondInteger, secondDecimal, secondString) { // Declare a variable named 'firstInteger' and initialize with integer value 4. const firstInteger = 4; // Declare a variable named 'firstDecimal' and initialize with floating-point value 4.0. const firstDecimal = 4.0; // Declare a variable named 'firstString' and initialize with the string "HackerRank". const firstString = 'HackerRank '; // Write code that uses console.log to print the sum of the 'firstInteger' and 'secondInteger' (converted to a Number type) on a new line. console.log(firstInteger + Number(secondInteger)); // Write code that uses console.log to print the sum of 'firstDecimal' and 'secondDecimal' (converted to a Number type) on a new line. console.log(firstDecimal + Number(secondDecimal)); // Write code that uses console.log to print the concatenation of 'firstString' and 'secondString' on a new line. The variable 'firstString' must be printed first. console.log(firstString + secondString); }
                                                                                                      Helper function for reverseIndex .
                                                                                                      javascriptdot imgLines of Code : 13dot imgno licencesLicense : No License
                                                                                                      copy iconCopy
                                                                                                      
                                                                                                                                          function reverseStringHalfIndex(str) { let strArr = str.split(''); let len = strArr.length; let halfIndex = Math.floor(len / 2) - 1; let tmp = []; for (var i = 0; i <= halfIndex; i++) { tmp = strArr[len - i - 1]; strArr[len - i - 1] = strArr[i]; // So for the first iteration I am doing str[len - 1] = str[0] strArr[i] = tmp; } return strArr.join(''); }
                                                                                                      Reverse an array
                                                                                                      javascriptdot imgLines of Code : 8dot imgno licencesLicense : No License
                                                                                                      copy iconCopy
                                                                                                      
                                                                                                                                          function inPlaceReverse(arr) { var i = 0; while (i < arr.length - 1 ) { arr.splice(i, 0, arr.pop()); i++; } return arr; }
                                                                                                      Community Discussions

                                                                                                      Trending Discussions on interviews

                                                                                                      How to find most common words from specific rows and column and list how often it occurs at data.csv?
                                                                                                      chevron right
                                                                                                      How to calculate cumulative sums in MySQL
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                                                                                                      QUESTION

                                                                                                      How to find most common words from specific rows and column and list how often it occurs at data.csv?
                                                                                                      Asked 2022-Mar-03 at 20:14

                                                                                                      I want to get 20 most common words from the descriptions of top 10 longest movies from data.csv, by using Python. So far, I got top 10 longest movies, however I am unable to get most common words from those specific movies, my code just gives most common words from whole data.csv itself. I tried Counter, Pandas, Numpy, Mathlib, but I have no idea how to make Python look exactly for most common words in the specific rows and column (description of movies) of the data table

                                                                                                      My code:

                                                                                                      import pandas as pd
                                                                                                      import numpy as np
                                                                                                      df = pd.read_csv("data.csv")
                                                                                                      small_df = df[['title','duration_min','description']]
                                                                                                      result_time = small_df.sort_values('duration_min', ascending=False)
                                                                                                      print("TOP 10 LONGEST: ")
                                                                                                      print(result_time.head(n=10))
                                                                                                      
                                                                                                      most_common = pd.Series(' '.join(result_time['description']).lower().split()).value_counts()[:20]
                                                                                                      print("20 Most common words from TOP 10 longest movies: ")
                                                                                                      print(most_common)
                                                                                                      

                                                                                                      My output:

                                                                                                      TOP 10 LONGEST: 
                                                                                                                                   title  duration_min                                        description
                                                                                                      6840        The School of Mischief         253.0  A high school teacher volunteers to transform ...
                                                                                                      4482                No Longer kids         237.0  Hoping to prevent their father from skipping t...
                                                                                                      3687            Lock Your Girls In         233.0  A widower believes he must marry off his three...
                                                                                                      5100               Raya and Sakina         230.0  When robberies and murders targeting women swe...
                                                                                                      5367                        Sangam         228.0  Returning home from war after being assumed de...
                                                                                                      3514                        Lagaan         224.0  In 1890s India, an arrogant British commander ...
                                                                                                      3190                  Jodhaa Akbar         214.0  In 16th-century India, what begins as a strate...
                                                                                                      6497                  The Irishman         209.0  Hit man Frank Sheeran looks back at the secret...
                                                                                                      3277      Kabhi Khushi Kabhie Gham         209.0  Years after his father disowns his adopted bro...
                                                                                                      4476  No Direction Home: Bob Dylan         208.0  Featuring rare concert footage and interviews ...
                                                                                                      20 Most common words from TOP 10 longest movies: 
                                                                                                      a        10134
                                                                                                      the       7153
                                                                                                      to        5653
                                                                                                      and       5573
                                                                                                      of        4691
                                                                                                      in        3840
                                                                                                      his       3005
                                                                                                      with      1967
                                                                                                      her       1803
                                                                                                      an        1727
                                                                                                      for       1558
                                                                                                      on        1528
                                                                                                      their     1468
                                                                                                      when      1320
                                                                                                      this      1240
                                                                                                      from      1114
                                                                                                      as        1050
                                                                                                      is         988
                                                                                                      by         894
                                                                                                      after      865
                                                                                                      dtype: int64
                                                                                                      

                                                                                                      ANSWER

                                                                                                      Answered 2022-Mar-03 at 20:05

                                                                                                      You can select the first 10 rows of your dataframe with iloc[0:10].

                                                                                                      In this case, the solution would look like this, with the least modification to your existing code:

                                                                                                      import pandas as pd
                                                                                                      import numpy as np    
                                                                                                      df = pd.read_csv("data.csv")
                                                                                                      small_df = df[['title','duration_min','description']]
                                                                                                      result_time = small_df.sort_values('duration_min', ascending=False)
                                                                                                      print("TOP 10 LONGEST: ")
                                                                                                      print(result_time.head(n=10))
                                                                                                      
                                                                                                      most_common = pd.Series(' '.join(result_time.iloc[0:10]['description']).lower().split()).value_counts()[:20]
                                                                                                      print("20 Most common words from TOP 10 longest movies: ")
                                                                                                      print(most_common) 
                                                                                                      

                                                                                                      Source https://stackoverflow.com/questions/71343075

                                                                                                      QUESTION

                                                                                                      How to calculate cumulative sums in MySQL
                                                                                                      Asked 2022-Feb-21 at 07:02

                                                                                                      I am preparing for interviews and came across this question while practicing some SQL questions recently asked in Amazon. I could not find the table though, but the question is as follows:

                                                                                                      Find the cumulative sum of the top 10 most profitable products of the last 6 months for customers in Seattle.

                                                                                                      Does the approach to solving this type of query look correct? If not, what would be the best way to approach this problem?

                                                                                                      SELECT t.day,
                                                                                                             t.product_count,
                                                                                                             @running_total:=@running_total + t.product_count AS cumulative_sum
                                                                                                      FROM
                                                                                                      ( SELECT
                                                                                                        date(purchase_date) as day,
                                                                                                        count(product_id) as product_count
                                                                                                        FROM products
                                                                                                        where day > DATE_SUB(now(), INTERVAL 6 MONTH)
                                                                                                        AND customer_city = 'Seattle'
                                                                                                        GROUP BY day 
                                                                                                        ORDER BY product_count desc) t
                                                                                                      JOIN (SELECT @running_total:=0) r
                                                                                                      ORDER BY t.day
                                                                                                      LIMIT 10;
                                                                                                      

                                                                                                      ANSWER

                                                                                                      Answered 2022-Feb-21 at 07:02

                                                                                                      use this

                                                                                                      select day,product_count,
                                                                                                      sum(product_count) over (order by t.day ROWS UNBOUNDED PRECEDING) as cumulative_sum from (
                                                                                                      SELECT
                                                                                                        date(purchase_date) as day,
                                                                                                        count(product_id) as product_count
                                                                                                        FROM products
                                                                                                        where day > DATE_SUB(now(), INTERVAL 6 MONTH)
                                                                                                        AND customer_city = 'Seattle'
                                                                                                        GROUP BY day 
                                                                                                        ORDER BY product_count desc
                                                                                                      )t
                                                                                                      
                                                                                                      

                                                                                                      Source https://stackoverflow.com/questions/71202242

                                                                                                      QUESTION

                                                                                                      css grid relayout if element changes height
                                                                                                      Asked 2022-Jan-21 at 17:51

                                                                                                      I have a css grid layout that looks like this,

                                                                                                      When a box is clicked is grows in height to show information about the service. What I was hoping to be able to do was the "relayout" the grid so grid wrapped around the tallest item? Instead what I have is when I an item grows that row and it's children grow with it.

                                                                                                      What I was hoping for was if report writing was clicked it would grow and take up benchmarking space, benchmarking would move left and consultancy would wrap onto a new line?

                                                                                                      I am using tailwind so my HTML looks like this,

                                                                                                      
                                                                                                      
                                                                                                                                                          
                                                                                                                                  
                                                                                                                                                                      
                                                                                                                                  Report Writing
                                                                                                                                  DBC can provide reports on any activity for organisations wishing to change any aspect of its operation or strategic direction. David has written consultancy reports for many countries on library and information transformation projects, various institutions seeking to gain taught degree awarding powers (TDAP), leadership and management benchmarking exercises and organisational training provision.
                                                                                                      Notable recent examples include three reports written for for SCONUL to assess the strategic role of libraries and their leaders from the viewpoint of the universities’ senior management. This resulted in a publication for the commissioning body SCONUL to highlight the expectations of university leadership in the development of relevant library strategies. As a further phase of this work, Alison Allden and David Baker looked at the opportunities and transferable skills relating to international movement amongst library leaders.
                                                                                                      David has been working with the University of London since March 2020 to develop a new strategy for Senate House Library, along with the School of Advanced Study, Federal Member Institute Libraries and the University of London Worldwide. The initial report was accepted by the University in September 2020. In the light of this, David has created a major Library Transformation Programme (LTP). The work involved in developing and operationalising the report included surveys, benchmarking, focus groups and workshops. 
                                                                                                                          
                                                                                                                                              
                                                                                                                                  
                                                                                                                                                                      
                                                                                                                                  Mentoring
                                                                                                                                  David has experience of mentoring professionals from different sectors and academics reaching back to 1990. He is currently heavily involved in mentoring for CILIP Chartership, Certification and Fellowship status (the UK’s library and information association). The focus is on developing and nurturing staff while learning new ideas and approaches from other professionals. Mentoring can be undertaken in person, by email, by telephone or through online platforms. 
                                                                                                                          
                                                                                                                                              
                                                                                                                                  
                                                                                                                                                                      
                                                                                                                                  Workshops
                                                                                                                                  DBC has a wealth of experience in delivering face-to-face and online workshops for a range of projects and purposes. These can be for the purpose of stakeholder engagement, data gathering or as part of a communications strategy. Workshops, interviews and focus groups are carried out consistently and coherently to give maximum value of data and information gathered. This is done through agreed pro forma and protocols. 
                                                                                                                          
                                                                                                                                              
                                                                                                                                  
                                                                                                                                                                   
                                                                                                                                  Training
                                                                                                                                  The DBC team can offer project management and change management training using previous live (anonymised) projects. A DBC Associate is PRINCE2 trained.
                                                                                                      David Baker has delivered training in library management and information systems for senior leaders, library tecnhnicians and assistants for many countries of the world including, Slovenia, Ireland, Kuwait, Hungary, Germany and Portugal and has published several training guides, including on the subject of co-operative training in this area. He has also provided training and development for third world countries such as Ethiopia and Nigeria and has published a book on this. 
                                                                                                                          
                                                                                                                                              
                                                                                                                                  
                                                                                                                                                                   
                                                                                                                                  Benchmarking
                                                                                                                                  In 2022, DBC is undertaking a major international benchmarking exercise based on the Association of Commonwealth Universities (ACU) model. It is being led by David, Caroline (Librarian of the University of Queensland for academic and research libraries in the Australia and South Pacific region), Cliff and Lucy. The title is "Benchmarking Library, Information And Education Services: New Strategic Choices In Challenging Times". An Elsevier book publication bearing the same title will be published in early 2023. The benchmarking model can be adapted for any organisational purpose. 
                                                                                                                          
                                                                                                                                              
                                                                                                                                  
                                                                                                                                                                   
                                                                                                                                  Research Services
                                                                                                                                  DBC Associates worked in 2021 on a scoping study commissioned by Research Libraries UK (RLUK) and funded by the Arts and Humanities Research Council (AHRC). It resulted in a wealth of evidence of the role and potential of research libraries as partners and leaders of research, contributing to longer-term strategic development in the process. A range of research techniques were used for this and other consultancies such as surveys, benchmarking, focus groups and workshops. Two senior associates have research backgrounds and PhDs and the Director, Professor, has a substantial and significant track record in research and publishing. 
                                                                                                                          
                                                                                                                                              
                                                                                                                                  
                                                                                                                                                                   
                                                                                                                                  Organisation Development
                                                                                                                                  DBC has been developing strategic plans at organisational and pan-organisational levels for many years, not least through working in chief executive and governance roles as well as high-level consultancy work. We are accustomed to working with governing bodies, steering committees, task forces and other groupings to shape strategic direction and effect major organisational change as a result.
                                                                                                      Our biographical details demonstrate that we have developed, written and implemented many strategic plans, including for the Higher Education Statistics Agency (HESA) and the Joint Information Systems Committee (Jisc), as well as contributing to strategy development for the Society of College, National and University Libraries (SCONUL) and Research Libraries UK (RLUK). 
                                                                                                                          
                                                                                                                                              
                                                                                                                                  
                                                                                                                                                                   
                                                                                                                                  Consultancy
                                                                                                                                  DBC delivers high-quality consultancy projects in higher education both nationally and internationally, with a long-standing track record, especially in strategy development. Associates have a broad and deep knowledge of the field. 
                                                                                                                          
                                                                                                                                              
                                                                                                      

                                                                                                      Here is a codepen of my current solution

                                                                                                      ANSWER

                                                                                                      Answered 2022-Jan-21 at 17:51

                                                                                                      A couple of things.

                                                                                                      You can make the clicked item span two rows by setting grid-row: span 2 This will have the effect of 'pushing' other grid items around.

                                                                                                      In the JS you had a call to remove which I think should have been removeClass

                                                                                                      Here's a (slightly messy) SO snippet created from your codepen:

                                                                                                      
                                                                                                      
                                                                                                        
                                                                                                      
                                                                                                        
                                                                                                        
                                                                                                      
                                                                                                        
                                                                                                      
                                                                                                        
                                                                                                      
                                                                                                      
                                                                                                        
                                                                                                          CodePen - A Pen by Simon Ainley
                                                                                                      
                                                                                                      
                                                                                                          
                                                                                                      
                                                                                                          
                                                                                                      
                                                                                                          
                                                                                                      
                                                                                                      
                                                                                                      
                                                                                                          
                                                                                                      
                                                                                                      
                                                                                                          
                                                                                                        
                                                                                                        
                                                                                                      
                                                                                                          
                                                                                                            
                                                                                                               
                                                                                                            Report Writing
                                                                                                            DBC can provide reports on any activity for organisations wishing to change any aspect of its operation or strategic direction. David Baker has written consultancy reports for many countries on library and information transformation projects, various
                                                                                                              institutions seeking to gain taught degree awarding powers (TDAP), leadership and management benchmarking exercises and organisational training provision. Notable recent examples include three reports written for for SCONUL to assess the strategic
                                                                                                              role of libraries and their leaders from the viewpoint of the universities’ senior management. This resulted in a publication for the commissioning body SCONUL to highlight the expectations of university leadership in the development of relevant
                                                                                                              library strategies. As a further phase of this work, Alison Allden and David Baker looked at the opportunities and transferable skills relating to international movement amongst library leaders. David Baker has been working with the University
                                                                                                              of London since March 2020 to develop a new strategy for Senate House Library, along with the School of Advanced Study, Federal Member Institute Libraries and the University of London Worldwide. The initial report was accepted by the University
                                                                                                              in September 2020. In the light of this, David has created a major Library Transformation Programme (LTP). The work involved in developing and operationalising the report included surveys, benchmarking, focus groups and workshops.
                                                                                                          
                                                                                                          
                                                                                                            
                                                                                                               
                                                                                                            Mentoring
                                                                                                            Professor David Baker has experience of mentoring professionals from different sectors and academics reaching back to 1990. He is currently heavily involved in mentoring for CILIP Chartership, Certification and Fellowship status (the UK’s library
                                                                                                              and information association). The focus is on developing and nurturing staff while learning new ideas and approaches from other professionals. Mentoring can be undertaken in person, by email, by telephone or through online platforms.
                                                                                                          
                                                                                                          
                                                                                                            
                                                                                                               
                                                                                                            Workshops
                                                                                                            DBC has a wealth of experience in delivering face-to-face and online workshops for a range of projects and purposes. These can be for the purpose of stakeholder engagement, data gathering or as part of a communications strategy. Workshops, interviews
                                                                                                              and focus groups are carried out consistently and coherently to give maximum value of data and information gathered. This is done through agreed pro forma and protocols.
                                                                                                          
                                                                                                          
                                                                                                            
                                                                                                               
                                                                                                            Training
                                                                                                            The DBC team can offer project management and change management training using previous live (anonymised) projects. A DBC Associate is PRINCE2 trained. David Baker has delivered training in library management and information systems for senior leaders,
                                                                                                              library tecnhnicians and assistants for many countries of the world including, Slovenia, Ireland, Kuwait, Hungary, Germany and Portugal and has published several training guides, including on the subject of co-operative training in this area.
                                                                                                              He has also provided training and development for third world countries such as Ethiopia and Nigeria and has published a book on this.
                                                                                                          
                                                                                                          
                                                                                                            
                                                                                                               
                                                                                                            Benchmarking
                                                                                                            In 2022, DBC is undertaking a major international benchmarking exercise based on the Association of Commonwealth Universities (ACU) model. It is being led by David Baker, Caroline Williams (Librarian of the University of Queensland for academic
                                                                                                              and research libraries in the Australia and South Pacific region), Cliff Wragg and Lucy Ellis. The title is "Benchmarking Library, Information And Education Services: New Strategic Choices In Challenging Times". An Elsevier book publication bearing
                                                                                                              the same title will be published in early 2023. The benchmarking model can be adapted for any organisational purpose.
                                                                                                          
                                                                                                          
                                                                                                            
                                                                                                               
                                                                                                            Research Services
                                                                                                            DBC Associates worked in 2021 on a scoping study commissioned by Research Libraries UK (RLUK) and funded by the Arts and Humanities Research Council (AHRC). It resulted in a wealth of evidence of the role and potential of research libraries as partners
                                                                                                              and leaders of research, contributing to longer-term strategic development in the process. A range of research techniques were used for this and other consultancies such as surveys, benchmarking, focus groups and workshops. Two senior associates
                                                                                                              have research backgrounds and PhDs and the Director, Professor Baker, has a substantial and significant track record in research and publishing.
                                                                                                          
                                                                                                          
                                                                                                            
                                                                                                               
                                                                                                            Organisation Development
                                                                                                            DBC has been developing strategic plans at organisational and pan-organisational levels for many years, not least through working in chief executive and governance roles as well as high-level consultancy work. We are accustomed to working with governing
                                                                                                              bodies, steering committees, task forces and other groupings to shape strategic direction and effect major organisational change as a result. Our biographical details demonstrate that we have developed, written and implemented many strategic plans,
                                                                                                              including for the Higher Education Statistics Agency (HESA) and the Joint Information Systems Committee (Jisc), as well as contributing to strategy development for the Society of College, National and University Libraries (SCONUL) and Research
                                                                                                              Libraries UK (RLUK).
                                                                                                          
                                                                                                          
                                                                                                            
                                                                                                               
                                                                                                            Consultancy
                                                                                                            DBC delivers high-quality consultancy projects in higher education both nationally and internationally, with a long-standing track record, especially in strategy development. Associates have a broad and deep knowledge of the field.
                                                                                                          
                                                                                                        
                                                                                                        
                                                                                                        

                                                                                                      Source https://stackoverflow.com/questions/70804213

                                                                                                      QUESTION

                                                                                                      Is it acceptable to pass this amount of JSON information as a parameter to a C# COM DLL method?
                                                                                                      Asked 2022-Jan-21 at 05:06

                                                                                                      This question is an extension to this one (Is there a technical reason why it would be better for the COM DLL to delete the passed in temporary JSON when it is finished with it?) where it was suggested I pass JSON content as a BSTR to my C# COM DLL.

                                                                                                      Here is an example of the type of data being passed:

                                                                                                      {
                                                                                                        "BibleReading": "Bible Reading",
                                                                                                        "BibleReadingMain": "Bible Reading (Main)",
                                                                                                        "BibleReadingAux": "Bible Reading (Aux)",
                                                                                                        "InitialCall": "Initial Call",
                                                                                                        "InitialCallMain": "Initial Call (Main)",
                                                                                                        "InitialCallAux": "Initial Call (Aux)",
                                                                                                        "ReturnVisit": "Return Visit",
                                                                                                        "ReturnVisitMain": "Return Visit (Main)",
                                                                                                        "ReturnVisitAux": "Return Visit (Aux)",
                                                                                                        "BibleStudy": "Bible Study",
                                                                                                        "BibleStudyMain": "Bible Study (Main)",
                                                                                                        "BibleStudyAux": "Bible Study (Aux)",
                                                                                                        "Talk": "Talk",
                                                                                                        "TalkMain": "Talk (Main)",
                                                                                                        "TalkAux": "Talk (Aux)",
                                                                                                        "Assistant": "Assistant",
                                                                                                        "QuestionsAndAnswers": "Questions and Answers",
                                                                                                        "DiscussionWithVideo": "Discussion with Video",
                                                                                                        "SampleConversation": "Sample Conversation",
                                                                                                        "InitialCallVideo": "Initial Call Video",
                                                                                                        "ReturnVisitVideo": "Return Visit Video",
                                                                                                        "Video": "Video",
                                                                                                        "Presentations": "Prepare This Month’s Presentations",
                                                                                                        "SpiritualGems": "Spiritual Gems",
                                                                                                      }
                                                                                                      

                                                                                                      I want to extent this JSON to include 15 more items:

                                                                                                      {
                                                                                                        "BibleReading": "Bible Reading",
                                                                                                        "BibleReadingMain": "Bible Reading (Main)",
                                                                                                        "BibleReadingAux": "Bible Reading (Aux)",
                                                                                                        "InitialCall": "Initial Call",
                                                                                                        "InitialCallMain": "Initial Call (Main)",
                                                                                                        "InitialCallAux": "Initial Call (Aux)",
                                                                                                        "ReturnVisit": "Return Visit",
                                                                                                        "ReturnVisitMain": "Return Visit (Main)",
                                                                                                        "ReturnVisitAux": "Return Visit (Aux)",
                                                                                                        "BibleStudy": "Bible Study",
                                                                                                        "BibleStudyMain": "Bible Study (Main)",
                                                                                                        "BibleStudyAux": "Bible Study (Aux)",
                                                                                                        "Talk": "Talk",
                                                                                                        "TalkMain": "Talk (Main)",
                                                                                                        "TalkAux": "Talk (Aux)",
                                                                                                        "Assistant": "Assistant",
                                                                                                        "QuestionsAndAnswers": "Questions and Answers",
                                                                                                        "DiscussionWithVideo": "Discussion with Video",
                                                                                                        "SampleConversation": "Sample Conversation",
                                                                                                        "InitialCallVideo": "Initial Call Video",
                                                                                                        "ReturnVisitVideo": "Return Visit Video",
                                                                                                        "Video": "Video",
                                                                                                        "Presentations": "Prepare This Month’s Presentations",
                                                                                                        "SpiritualGems": "Spiritual Gems",
                                                                                                        "Discuss": "Discussion",
                                                                                                        "DiscussDemos": "Discussion with Demonstration(s)",
                                                                                                        "DiscussDemosInterviews": "Discussion with Demonstration(s) and Interview(s)",
                                                                                                        "DiscussInterviews": "Discussion with Interview(s)",
                                                                                                        "DiscussVideo": "Discussion with Video",
                                                                                                        "DiscussVideos": "Discussion with Videos",
                                                                                                        "QuestionAnswer": "Questions and Answers",
                                                                                                        "QuestionAnswerDemos": "Questions and Answers with Demonstration(s)",
                                                                                                        "QuestionAnswerDemosInterviews": "Questions and Answers with Demonstration(s) and Interview(s)",
                                                                                                        "QuestionAnswerInterviews": "Questions and Answers with Interview(s)",
                                                                                                        "TalkDemos": "Talk with Demonstration(s)",
                                                                                                        "TalkDemosInterviews": "Talk with Demonstration(s) and Interview(s)",
                                                                                                        "TalkInterviews": "Talk with Interview(s)",
                                                                                                        "TalkVideo": "Talk with Video",
                                                                                                        "TalkVideos": "Talk with Videos"
                                                                                                      }
                                                                                                      

                                                                                                      Is it still going to be acceptable to pass this amount of information as a BSTR to my COM API method? I assume there is some kind of limit as to how much data can be passed.

                                                                                                      Thanks for confirming.

                                                                                                      ANSWER

                                                                                                      Answered 2022-Jan-21 at 05:06

                                                                                                      You could always write a test program if you don't know the answer. Here's one that verifies 1 megabyte BSTR which is way more than your example. You could change the amount allocated to whatever you want. At some point it will break.

                                                                                                      #include 
                                                                                                      #include 
                                                                                                      #include 
                                                                                                      #include 
                                                                                                      #include 
                                                                                                      
                                                                                                      using namespace std;
                                                                                                      
                                                                                                      int main(int argc, char* argv)
                                                                                                      {
                                                                                                          const size_t MEGABYTEPLUSONE = 1024 * 1024 + 1;
                                                                                                          auto pChar = std::make_unique(MEGABYTEPLUSONE);
                                                                                                          for (size_t i = 0; i < MEGABYTEPLUSONE - 2; ++i)
                                                                                                              pChar[i] = (i % 26) + 65;
                                                                                                      
                                                                                                          pChar[MEGABYTEPLUSONE - 1] = 0;
                                                                                                          
                                                                                                          _bstr_t bstr(pChar.get());
                                                                                                          int len = bstr.length();
                                                                                                          wcout << L"Length of BSTR: " << len << endl;
                                                                                                          const WCHAR* const pwcStart = bstr.GetBSTR();
                                                                                                          for (const WCHAR* pwc = pwcStart;  pwc < pwcStart + 52; ++pwc)
                                                                                                              wcout << *pwc;
                                                                                                          wcout << endl;
                                                                                                          
                                                                                                          return 0;
                                                                                                      }
                                                                                                      

                                                                                                      Source https://stackoverflow.com/questions/70778361

                                                                                                      QUESTION

                                                                                                      matrix maximum diagonal pattern match
                                                                                                      Asked 2022-Jan-16 at 03:00

                                                                                                      I came across a problem while doing technical interviews for matrix pattern matching. I was able to solve the problem using brute force but wonder if there is a more efficient solution as I only got about half credit for efficiency. Thank you in advanced for your input.

                                                                                                      Given a matrix containing the numbers 0, 1, 2 and the pattern [1, 2, 0, 2, 0, 2, 0] find the max length of the matching pattern starting from any point in the matrix but can only travel diagonally.

                                                                                                      here is an example of where we would expect the function to return 12.

                                                                                                      here is where we would expect the function to return 7.

                                                                                                      and empty matrix should return 0.

                                                                                                      Here is my code, like I said previously it does work and passed all the tests but I got docked points.

                                                                                                      def max_diagonal_match(matrix) -> int:
                                                                                                          max_len = 0
                                                                                                          pattern_match = [1, 2, 0, 2, 0, 2, 0]
                                                                                                          for i in range(0, len(matrix)):
                                                                                                              for j in range(0, len(matrix[i])):
                                                                                                                  max_len = max(
                                                                                                                      max_len,
                                                                                                                      find_i_minus_j_minus(matrix, i, j, pattern_match, 7),
                                                                                                                      find_i_minus_j_plus(matrix, i, j, pattern_match, 7),
                                                                                                                      find_i_plus_j_minus(matrix, i, j, pattern_match, 7),
                                                                                                                      find_i_plus_j_plus(matrix, i, j, pattern_match, 7)
                                                                                                                  )
                                                                                                          return max_len
                                                                                                      
                                                                                                      
                                                                                                      # i-, j-
                                                                                                      def find_i_minus_j_minus(matrix, i, j, pattern, mod):
                                                                                                          count = 0
                                                                                                          while i >= 0 and j >= 0:
                                                                                                              if matrix[i][j] != pattern[count % mod]:
                                                                                                                  return count
                                                                                                              count += 1
                                                                                                              i -= 1
                                                                                                              j -= 1
                                                                                                          return count
                                                                                                      
                                                                                                      
                                                                                                      # i-, j+
                                                                                                      def find_i_minus_j_plus(matrix, i, j, pattern, mod):
                                                                                                          count = 0
                                                                                                          while i >= 0 and j < len(matrix[i]):
                                                                                                              if matrix[i][j] != pattern[count % mod]:
                                                                                                                  return count
                                                                                                              count += 1
                                                                                                              i -= 1
                                                                                                              j += 1
                                                                                                          return count
                                                                                                      
                                                                                                      
                                                                                                      # i+, j-
                                                                                                      def find_i_plus_j_minus(matrix, i, j, pattern, mod):
                                                                                                          count = 0
                                                                                                          while i < len(matrix) and j >= 0:
                                                                                                              if matrix[i][j] != pattern[count % mod]:
                                                                                                                  return count
                                                                                                              count += 1
                                                                                                              i += 1
                                                                                                              j -= 1
                                                                                                          return count
                                                                                                      
                                                                                                      
                                                                                                      # j+, i+
                                                                                                      def find_i_plus_j_plus(matrix, i, j, pattern, mod):
                                                                                                          count = 0
                                                                                                          while i < len(matrix) and j < len(matrix[i]):
                                                                                                              if matrix[i][j] != pattern[count % mod]:
                                                                                                                  return count
                                                                                                              count += 1
                                                                                                              i += 1
                                                                                                              j += 1
                                                                                                          return count
                                                                                                      
                                                                                                      

                                                                                                      Many thanks for taking the time to read and any input is much appreciated.

                                                                                                      ANSWER

                                                                                                      Answered 2022-Jan-15 at 22:48

                                                                                                      Building on the accept answer to this question. It's based on a few key ideas:

                                                                                                      You can use np.diagonal over a range to slice every possible diagonal out of the array. You can flip the array around to make sure you get all angles.

                                                                                                      You can tile the pattern to make sure it's at least as long or longer than the largest diagonal.

                                                                                                      You can zip the diagonal and the pattern and the extra values in the pattern will be dropped automatically due to the way zip works.

                                                                                                      Then you can use takewhile on the zipped (diag,pattern) to check how many consecutive matches there are len(set(x))==1.

                                                                                                      If you do this for all the possible combinations and take the max, you should have your answer.

                                                                                                      import numpy as np
                                                                                                      from math import ceil
                                                                                                      from itertools import takewhile
                                                                                                      
                                                                                                      def max_sequence(arr):
                                                                                                          solns = []
                                                                                                          i = arr.shape[0]
                                                                                                          for x in range(-i, i+1):
                                                                                                              values = arr.diagonal(x)
                                                                                                              N = len(values)
                                                                                                              possibles = np.where(values == pattern[0])[0]
                                                                                                              for p in possibles:
                                                                                                                  check = values[p:p+N]
                                                                                                                  m = len(list(takewhile(lambda x:len(set(x))==1, zip(pattern,check))))
                                                                                                                  solns.append(m)
                                                                                                          return max(solns)
                                                                                                      
                                                                                                      def find_longest(arr):
                                                                                                          if len(arr)>0:
                                                                                                              return max([max_sequence(x) for x in [arr, np.fliplr(arr), np.flipud(arr), arr[::-1]]])
                                                                                                          else:
                                                                                                              return 0
                                                                                                      
                                                                                                      arr = np.array([
                                                                                                          [1,0,2,1,1,1,1,0,0,0,0,0,0],
                                                                                                          [1,2,2,1,1,1,1,0,0,0,0,0,0],
                                                                                                          [1,0,0,1,1,1,1,0,0,0,0,0,0],
                                                                                                          [1,0,0,2,1,1,1,0,0,0,0,0,0],
                                                                                                          [1,0,0,2,0,1,1,0,0,0,0,0,0],
                                                                                                          [1,0,0,1,1,2,1,0,0,0,0,0,0],
                                                                                                          [1,0,0,1,1,1,0,0,0,0,0,0,0],
                                                                                                          [0,0,0,0,0,0,0,1,0,0,0,0,0],
                                                                                                          [0,0,0,0,0,0,0,0,2,0,0,0,0],
                                                                                                          [0,0,0,0,0,0,0,0,0,0,0,0,0],
                                                                                                          [0,0,0,0,0,0,0,0,0,0,2,0,0],
                                                                                                          [0,0,0,0,0,0,0,0,0,0,0,0,0],
                                                                                                      ])
                                                                                                      
                                                                                                      arr1 = np.array([
                                                                                                          [1,0,2,1,1,1,1],
                                                                                                          [1,2,2,1,1,1,1],
                                                                                                          [1,0,0,1,1,1,1],
                                                                                                          [1,0,0,2,1,1,1],
                                                                                                          [1,0,0,2,0,1,1],
                                                                                                          [1,0,0,1,1,2,1],
                                                                                                          [1,0,0,1,1,1,0]
                                                                                                      ])
                                                                                                      
                                                                                                      arr2 = np.array([])
                                                                                                      
                                                                                                      pattern = [1, 2, 0, 2, 0, 2, 0]
                                                                                                      # Make sure pattern repeats longer than the max diagonal
                                                                                                      pattern = np.tile(pattern,ceil(arr.shape[1] / len(pattern)))
                                                                                                      
                                                                                                      for a in [arr, arr1, arr2]:
                                                                                                          print(find_longest(a))
                                                                                                      

                                                                                                      Output

                                                                                                      12
                                                                                                      7
                                                                                                      0
                                                                                                      

                                                                                                      Source https://stackoverflow.com/questions/70725603

                                                                                                      QUESTION

                                                                                                      Cleaner/Simpler way to check if content in array has a value greater than 0
                                                                                                      Asked 2021-Dec-23 at 20:57

                                                                                                      I have this array that combines different data and then if the array contains true, then I'm showing some text. So for example, even if video is the only one that has a length greater than 0 (true), and the rest are false, the text 'Click to go to list' will be still shown. Is there a clean/simplier way of doing this check?

                                                                                                      const relatedLists = [
                                                                                                       videos.length > 0, // returns true or false
                                                                                                       interviews.length > 0,
                                                                                                       pdf.length > 0,
                                                                                                       audio.length > 0,
                                                                                                       book.length > 0,
                                                                                                      ];
                                                                                                      
                                                                                                      Link={
                                                                                                       relatedList.includes(true) ? 'Click to go to list' : ''
                                                                                                      }
                                                                                                      

                                                                                                      ANSWER

                                                                                                      Answered 2021-Dec-23 at 20:38

                                                                                                      You can check to see if atleast one length property in the array is truthy

                                                                                                      Link={relatedList.some(v => !!v.length) ? 'Click to go to list' : ''}
                                                                                                      

                                                                                                      Source https://stackoverflow.com/questions/70465984

                                                                                                      QUESTION

                                                                                                      How to combine columns, group them then get a total count?
                                                                                                      Asked 2021-Dec-08 at 21:09

                                                                                                      Below is a table that has candidate_id, two interviews they attended with the interviewer's name, and results for each interview.

                                                                                                      candidate_id interview_1 interview_2 result_1 result_2 1 Interviewer_A Interviewer_B Pass Pass 2 Interviewer_C Interviewer_D Pass Reject

                                                                                                      I need help to combine column interview_1 and interview_2 into one column, and count how many pass and reject each interviewer gave to the candidate, the result I expected to see as below:

                                                                                                      interviewer_name pass_count reject_count Interviewer_A 1 0 Interviewer_B 1 0 Interviewer_C 1 0 Interviewer_D 0 1

                                                                                                      SQL or Python either would work for me! Much appreciated!

                                                                                                      ANSWER

                                                                                                      Answered 2021-Dec-08 at 21:00

                                                                                                      In SQL Server, it becomes a small matter for a CROSS APPLY

                                                                                                      Example

                                                                                                      Select [candidate_id]
                                                                                                            ,B.[Interview_name]
                                                                                                            ,pass_count   = case when result='Pass' then 1 else 0 end
                                                                                                            ,reject_count = case when result='Pass' then 0 else 1 end
                                                                                                       From YourTable A
                                                                                                       Cross Apply ( values ([interview_1],[result_1])
                                                                                                                           ,([interview_2],[result_2])
                                                                                                                   ) B(Interview_name,result)
                                                                                                      

                                                                                                      Results

                                                                                                      candidate_id    Interview_name  pass_count  reject_count
                                                                                                      1               Interviewer_A   1           0
                                                                                                      1               Interviewer_B   1           0
                                                                                                      2               Interviewer_C   1           0
                                                                                                      2               Interviewer_D   0           1
                                                                                                      

                                                                                                      Source https://stackoverflow.com/questions/70281329

                                                                                                      QUESTION

                                                                                                      formulate capture groups for inconsistently present substrings
                                                                                                      Asked 2021-Nov-27 at 15:09

                                                                                                      I have transcriptions of interviews that are partly irregularly formed:

                                                                                                      tst <- c("In: ja COOL;  #00:04:24-6#  ",           
                                                                                                               "  in den vier, FÜNF wochen, #00:04:57-8# ",
                                                                                                               "In: jah,  #00:02:07-8# ",
                                                                                                               "In:     [ja; ] #00:03:25-5# [ja; ] #00:03:26-1#",
                                                                                                               "    also jA:h; #00:03:16-6# (1.1)",
                                                                                                               "Bz:        [E::hm;    ]  #00:03:51-4#  (3.0)  ",
                                                                                                               "Bz:    [mhmh,      ]",
                                                                                                               "  in den bilLIE da war;")
                                                                                                      

                                                                                                      What I need to do is structure this data by extracting its key elements into columns of a dataframe. There are four such key elements:

                                                                                                      • Rolein interview: interviewee or interviewer
                                                                                                      • Utterance: the interview partners' speech
                                                                                                      • Timestampindicated by # to both ends
                                                                                                      • Gap indicated by decimal number in brackets

                                                                                                      The problem is that both Timestamp and Gapare inconsistently provided. While I can make the last capture group for Gap optional, those strings that have neither Timestamp nor Gapare not rendered correctly:

                                                                                                      I'm using extract from tidyr for the extraction:

                                                                                                      library(tidyr)
                                                                                                      data.frame(tst) %>%
                                                                                                        extract(col = tst,
                                                                                                                into = c("Role", "Utterance", "Timestamp", "Gap"),
                                                                                                                regex = "^(\\w{2}:\\s|\\s+)([\\S\\s]+?)\\s*#([^#]+)?#\\s*(\\([0-9.]+\\))?\\s*")
                                                                                                        Role                 Utterance  Timestamp   Gap
                                                                                                      1 In:                   ja COOL; 00:04:24-6      
                                                                                                      2      in den vier, FÜNF wochen, 00:04:57-8      
                                                                                                      3 In:                       jah, 00:02:07-8      
                                                                                                      4 In:                     [ja; ] 00:03:25-5      
                                                                                                      5                     also jA:h; 00:03:16-6 (1.1)
                                                                                                      6 Bz:               [E::hm;    ] 00:03:51-4 (3.0)
                                                                                                      7                                
                                                                                                      8                                
                                                                                                      

                                                                                                      How can the regex be refined so that I get this desired output:

                                                                                                        Role                 Utterance  Timestamp   Gap
                                                                                                      1 In:                   ja COOL; 00:04:24-6      
                                                                                                      2      in den vier, FÜNF wochen, 00:04:57-8      
                                                                                                      3 In:                       jah, 00:02:07-8      
                                                                                                      4 In:                     [ja; ] 00:03:25-5      
                                                                                                      5                     also jA:h; 00:03:16-6 (1.1)
                                                                                                      6 Bz:               [E::hm;    ] 00:03:51-4 (3.0)
                                                                                                      7 Bz:              [mhmh,      ]
                                                                                                      8          in den bilLIE da war;
                                                                                                      

                                                                                                      ANSWER

                                                                                                      Answered 2021-Nov-27 at 13:41

                                                                                                      You could update your pattern to use your 4 capture groups, and make the last part optional by optionally matching the 3rd group and then the 4th group and assert the end of the string:

                                                                                                      library(tidyr)
                                                                                                      
                                                                                                      tst <- c("In: ja COOL;  #00:04:24-6#  ",           
                                                                                                               "  in den vier, FÜNF wochen, #00:04:57-8# ",
                                                                                                               "In: jah,  #00:02:07-8# ",
                                                                                                               "In:     [ja; ] #00:03:25-5# [ja; ] #00:03:26-1#",
                                                                                                               "    also jA:h; #00:03:16-6# (1.1)",
                                                                                                               "Bz:        [E::hm;    ]  #00:03:51-4#  (3.0)  ",
                                                                                                               "Bz:    [mhmh,      ]",
                                                                                                               "  in den bilLIE da war;")     
                                                                                                      
                                                                                                      data.frame(tst) %>%
                                                                                                        extract(col = tst,
                                                                                                                into = c("Role", "Utterance", "Timestamp", "Gap"),
                                                                                                                regex = "^(\\w{2}:\\s|\\s+)([\\s\\S]*?)(?:\\s*#([^#]+)(?:#\\s*(\\([0-9.]+\\))?\\s*)?)?$")
                                                                                                      

                                                                                                      Output

                                                                                                        Role                      Utterance  Timestamp   Gap
                                                                                                      1 In:                        ja COOL; 00:04:24-6      
                                                                                                      2           in den vier, FÜNF wochen, 00:04:57-8      
                                                                                                      3 In:                            jah, 00:02:07-8      
                                                                                                      4 In:      [ja; ] #00:03:25-5# [ja; ] 00:03:26-1      
                                                                                                      5                          also jA:h; 00:03:16-6 (1.1)
                                                                                                      6 Bz:                    [E::hm;    ] 00:03:51-4 (3.0)
                                                                                                      7 Bz:                   [mhmh,      ]                 
                                                                                                      8               in den bilLIE da war; 
                                                                                                      

                                                                                                      Source https://stackoverflow.com/questions/70134684

                                                                                                      QUESTION

                                                                                                      SQL Query to find people who have interviews in different rooms inany days
                                                                                                      Asked 2021-Nov-25 at 11:12
                                                                                                      Name Date Room Jerry 2-2-21 D Sam 2-2-21 A Sarah 2-2-21 A Will 3-2-21 B Sam 4-3-21 D Will 2-2-21 B Jerry 2-2-21 D

                                                                                                      Hello, (apologies for my bad English) I made the previous table I'm a new at SQL and wondering to how can I make a query that will return the names of the people who had interviews in different rooms regardless of the day so will would not come up since he had interviews in the same rooms

                                                                                                      my approach

                                                                                                      SELECT name
                                                                                                      FROM worktable
                                                                                                      DISTINCT room > 2
                                                                                                      

                                                                                                      I don't' know what else to do thank you any help appreciated

                                                                                                      ANSWER

                                                                                                      Answered 2021-Nov-24 at 13:22

                                                                                                      In addition to @Beso answer I would also use DISTINCT in COUNT to be sure there are different rooms.

                                                                                                      SELECT name
                                                                                                      FROM worktable
                                                                                                      GROUP BY name
                                                                                                      HAVING COUNT(DISTINCT name) > 1
                                                                                                      

                                                                                                      In this case only Sam will be included in the select output.

                                                                                                      Source https://stackoverflow.com/questions/70096433

                                                                                                      QUESTION

                                                                                                      Complexity analysis for the permutations algorithm
                                                                                                      Asked 2021-Nov-14 at 15:07

                                                                                                      I'm trying to understand the time and space complexity of an algorithm for generating an array's permutations. Given a partially built permutation where k out of n elements are already selected, the algorithm selects element k+1 from the remaining n-k elements and calls itself to select the remaining n-k-1 elements:

                                                                                                      public static List> permutations(List A) {
                                                                                                          List> result = new ArrayList<>();
                                                                                                          permutations(A, 0, result);
                                                                                                          return result;
                                                                                                        }
                                                                                                      
                                                                                                        public static void permutations(List A, int start, List> result) {
                                                                                                          if(A.size()-1==start) {
                                                                                                            result.add(new ArrayList<>(A));
                                                                                                            return;
                                                                                                          }
                                                                                                      
                                                                                                          for (int i=start; i

                                                                                                      My thoughts are that in each call we swap the collection's elements 2n times, where n is the number of elements to permute, and make n recursive calls. So the running time seems to fit the recurrence relation T(n)=nT(n-1)+n=n[(n-1)T(n-2)+(n-1)]+n=...=n+n(n-1)+n(n-1)(n-2)+...+n!=n![1/(n-1)!+1/(n-2)!+...+1]=n!e, hence the time complexity is O(n!) and the space complexity is O(max(n!, n)), where n! is the total number of permutations and n is the height of the recursion tree.

                                                                                                      This problem is taken from the Elements of Programming Interviews book, and they're saying that the time complexity is O(n*n!) because "The number of function calls C(n)=1+nC(n-1) ... [which solves to] O(n!) ... [and] ... we do O(n) computation per call outside of the recursive calls".

                                                                                                      Which time complexity is correct?

                                                                                                      ANSWER

                                                                                                      Answered 2021-Nov-14 at 15:07

                                                                                                      The time complexity of this algorithm, counted by the number of basic operations performed, is Θ(n * n!). Think about the size of the result list when the algorithm terminates-- it contains n! permutations, each of length n, and we cannot create a list with n * n! total elements in less than that amount of time. The space complexity is the same, since the recursion stack only ever has O(n) calls at a time, so the size of the output list dominates the space complexity.

                                                                                                      If you count only the number of recursive calls to permutations(), the function is called O(n!) times, although this is usually not what is meant by 'time complexity' without further specification. In other words, you can generate all permutations in O(n!) time, as long as you don't read or write those permutations after they are generated.

                                                                                                      The part where your derivation of run-time breaks down is in the definition of T(n). If you define T(n) as 'the run-time of permutations(A, start) when the input, A, has length n', then you can not define it recursively in terms of T(n-1) or any other function of T(), because the length of the input in all recursive calls is n, the length of A.

                                                                                                      A more useful way to define T(n) is by specifying it as the run-time of permutations(A', start), when A' is any permutation of a fixed, initial array A, and A.length - start == n. It's easy to write the recurrence relation here:

                                                                                                      T(x) = x * T(x-1) + O(x)     if x > 1
                                                                                                      T(1) = A.length
                                                                                                      

                                                                                                      This takes into account the fact that the last recursive call, T(1), has to perform O(A.length) work to copy that array to the output, and this new recurrence gives the result from the textbook.

                                                                                                      Source https://stackoverflow.com/questions/69963601

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