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leetcode | Python & JAVA Solutions for Leetcode | Learning library

 by   qiyuangong Python Version: Current License: MIT

 by   qiyuangong Python Version: Current License: MIT

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kandi X-RAY | leetcode Summary

leetcode is a Python library typically used in Tutorial, Learning, Example Codes applications. leetcode has no vulnerabilities, it has a Permissive License and it has medium support. However leetcode has 6 bugs and it build file is not available. You can download it from GitHub.
Remember solutions are only solutions to given problems. If you want full study checklist for code & whiteboard interview, please turn to jwasham's coding-interview-university. Also, there are open source implementations for basic data structs and algorithms, such as Algorithms in Python and Algorithms in Java. I'm currently working on Analytics-Zoo - an unified Data Analytics and AI platform. Check it out, if you are interested in big data and deep learning.
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kandi-support Support

  • leetcode has a medium active ecosystem.
  • It has 3418 star(s) with 1001 fork(s). There are 107 watchers for this library.
  • It had no major release in the last 12 months.
  • There are 1 open issues and 5 have been closed. On average issues are closed in 109 days. There are 4 open pull requests and 0 closed requests.
  • It has a neutral sentiment in the developer community.
  • The latest version of leetcode is current.
leetcode Support
Best in #Learning
Average in #Learning
leetcode Support
Best in #Learning
Average in #Learning

quality kandi Quality

  • leetcode has 6 bugs (2 blocker, 0 critical, 2 major, 2 minor) and 1211 code smells.
leetcode Quality
Best in #Learning
Average in #Learning
leetcode Quality
Best in #Learning
Average in #Learning

securitySecurity

  • leetcode has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.
  • leetcode code analysis shows 0 unresolved vulnerabilities.
  • There are 1 security hotspots that need review.
leetcode Security
Best in #Learning
Average in #Learning
leetcode Security
Best in #Learning
Average in #Learning

license License

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

buildReuse

  • leetcode releases are not available. You will need to build from source code and install.
  • leetcode has no build file. You will be need to create the build yourself to build the component from source.
  • leetcode saves you 3748 person hours of effort in developing the same functionality from scratch.
  • It has 7999 lines of code, 683 functions and 488 files.
  • It has low code complexity. Code complexity directly impacts maintainability of the code.
leetcode Reuse
Best in #Learning
Average in #Learning
leetcode Reuse
Best in #Learning
Average in #Learning
Top functions reviewed by kandi - BETA

kandi has reviewed leetcode and discovered the below as its top functions. This is intended to give you an instant insight into leetcode implemented functionality, and help decide if they suit your requirements.

  • Generate palindromes
    • Permute a permutation
    • Get_permute function
  • Returns the partition of the value at the given position
    • Return True if head is palindrome
    • Recursively recursively recursively subroutine
  • Minimal XOR
    • Count the number of elements in x
  • Compute the maximum overlap between a set of points
    • Generate the GCD
  • Generate a list of unique permutations
    • Returns the number of islands in a grid
      • Return the index of the substring in haystack
        • Find the length of the ladder
          • Adds key to the list
            • Determines if two trees are similar
              • Return the value of key in the tree
                • Calculates the number of islands between two points
                  • Find circular array loop
                    • Returns the length of the longest sequence of nums
                      • Insert new intervals into intervals
                        • Computes the minimum optimal cost for a given customer cost
                          • Find the next permutation
                            • Reverse words
                              • Flatten a tree
                                • Reverse node in the list

                                  Get all kandi verified functions for this library.

                                  Get all kandi verified functions for this library.

                                  leetcode Key Features

                                  Python & JAVA Solutions for Leetcode

                                  Why is the output an empty list, when I call a function that skips nodes in a linked-list?

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                                  class Solution:
                                      def removeElements(self, head: ListNode, val: int) -> ListNode:
                                          curr = head
                                          dummy = ListNode(0, head)  # Link to head!
                                          prev = dummy
                                  
                                          while curr: 
                                              if curr.val == val:    
                                                  prev.next = curr.next  # This is all that needs to happen
                                              else: 
                                                  prev = curr
                                              curr = curr.next  # This should happen in both cases
                                              # The loop invariant is: prev.next == curr
                                  
                                          return dummy.next        
                                  
                                  class Solution:
                                      def removeElements(self, head: ListNode, val: int) -> ListNode:
                                          curr = head
                                          dummy = ListNode(0, head)
                                          prev = dummy
                                  
                                          def skipper(curr):
                                              while curr is not None and curr.val == val:
                                                  curr = curr.next
                                              return curr
                                          
                                          while curr: 
                                              if curr.val == val:    
                                                  prev.next = skipper(curr)
                                              else: 
                                                  prev = curr
                                              curr = prev.next
                                  
                                          return dummy.next
                                  
                                  class Solution:
                                      def removeElements(self, head: ListNode, val: int) -> ListNode:
                                          curr = head
                                          dummy = ListNode(0, head)  # Link to head!
                                          prev = dummy
                                  
                                          while curr: 
                                              if curr.val == val:    
                                                  prev.next = curr.next  # This is all that needs to happen
                                              else: 
                                                  prev = curr
                                              curr = curr.next  # This should happen in both cases
                                              # The loop invariant is: prev.next == curr
                                  
                                          return dummy.next        
                                  
                                  class Solution:
                                      def removeElements(self, head: ListNode, val: int) -> ListNode:
                                          curr = head
                                          dummy = ListNode(0, head)
                                          prev = dummy
                                  
                                          def skipper(curr):
                                              while curr is not None and curr.val == val:
                                                  curr = curr.next
                                              return curr
                                          
                                          while curr: 
                                              if curr.val == val:    
                                                  prev.next = skipper(curr)
                                              else: 
                                                  prev = curr
                                              curr = prev.next
                                  
                                          return dummy.next
                                  

                                  how can i check if a string contains another string throughout it

                                  copy iconCopydownload iconDownload
                                  import re
                                  l1 = ["heollo", "leetcode"]
                                  l2 = ["e", "oo"]
                                  patterns = ['.*'.join(list(item)) for item in l2]
                                  # patterns is ['e', 'o.*o']
                                  result = [item for item in l1 if all(
                                      [re.search(pattern, item) is not None for pattern in patterns]
                                  )]
                                  # ['heollo']
                                  

                                  Time and Space complexity of Palindrome Partitioning II

                                  copy iconCopydownload iconDownload
                                  T(0) = 1
                                  T(n) = O(n) + 2 * (sum of T(x) for x=0 to n-1)
                                  # and for simplicity, I will just use
                                  T(n) = n + 2 * (sum of T(x) for x=0 to n-1)
                                  
                                  T(1)  =   3 = 1 + 2 *  1  =  1 + 2 * (T(0))
                                  T(2)  =  10 = 2 + 2 *  4  =  2 + 2 * (T(0) + T(1))
                                  T(3)  =  31 = 3 + 2 * 14  =  3 + 2 * (T(0) + T(1) + T(2))
                                  T(4)  =  94 = 4 + 2 * 45  =  4 + 2 * (T(0) + T(1) + T(2) + T(3))
                                  
                                  T(n) = n + 2 * (sum of T(x) for x=0 to n-1)
                                  T(n) = n + 2 * (T(0) + T(1) + ... + T(n-1))
                                  T(n) = n + 2 * (T(0) + T(1) + ... + T(n-2)) + 2 * T(n-1)
                                  T(n) = 1 +  n-1 + 2 * (T(0) + T(1) + ... + T(n-2))  +  2 * T(n-1)
                                  
                                  # with
                                  T(n-1) = n-1 + 2 * (sum of T(x) for x=0 to n-2)
                                  T(n-1) = n-1 + 2 * (T(0) + T(1) + ... + T(n-2))
                                  
                                  # we can substitute:
                                  T(n) = 1 + T(n-1) + 2 * T(n-1)
                                  T(n) = 1 + 3 * T(n-1)
                                  
                                  S(0) = 1
                                  S(n) = 1 + S(n-1)
                                  
                                  T(0) = 1
                                  T(n) = O(n) + 2 * (sum of T(x) for x=0 to n-1)
                                  # and for simplicity, I will just use
                                  T(n) = n + 2 * (sum of T(x) for x=0 to n-1)
                                  
                                  T(1)  =   3 = 1 + 2 *  1  =  1 + 2 * (T(0))
                                  T(2)  =  10 = 2 + 2 *  4  =  2 + 2 * (T(0) + T(1))
                                  T(3)  =  31 = 3 + 2 * 14  =  3 + 2 * (T(0) + T(1) + T(2))
                                  T(4)  =  94 = 4 + 2 * 45  =  4 + 2 * (T(0) + T(1) + T(2) + T(3))
                                  
                                  T(n) = n + 2 * (sum of T(x) for x=0 to n-1)
                                  T(n) = n + 2 * (T(0) + T(1) + ... + T(n-1))
                                  T(n) = n + 2 * (T(0) + T(1) + ... + T(n-2)) + 2 * T(n-1)
                                  T(n) = 1 +  n-1 + 2 * (T(0) + T(1) + ... + T(n-2))  +  2 * T(n-1)
                                  
                                  # with
                                  T(n-1) = n-1 + 2 * (sum of T(x) for x=0 to n-2)
                                  T(n-1) = n-1 + 2 * (T(0) + T(1) + ... + T(n-2))
                                  
                                  # we can substitute:
                                  T(n) = 1 + T(n-1) + 2 * T(n-1)
                                  T(n) = 1 + 3 * T(n-1)
                                  
                                  S(0) = 1
                                  S(n) = 1 + S(n-1)
                                  
                                  T(0) = 1
                                  T(n) = O(n) + 2 * (sum of T(x) for x=0 to n-1)
                                  # and for simplicity, I will just use
                                  T(n) = n + 2 * (sum of T(x) for x=0 to n-1)
                                  
                                  T(1)  =   3 = 1 + 2 *  1  =  1 + 2 * (T(0))
                                  T(2)  =  10 = 2 + 2 *  4  =  2 + 2 * (T(0) + T(1))
                                  T(3)  =  31 = 3 + 2 * 14  =  3 + 2 * (T(0) + T(1) + T(2))
                                  T(4)  =  94 = 4 + 2 * 45  =  4 + 2 * (T(0) + T(1) + T(2) + T(3))
                                  
                                  T(n) = n + 2 * (sum of T(x) for x=0 to n-1)
                                  T(n) = n + 2 * (T(0) + T(1) + ... + T(n-1))
                                  T(n) = n + 2 * (T(0) + T(1) + ... + T(n-2)) + 2 * T(n-1)
                                  T(n) = 1 +  n-1 + 2 * (T(0) + T(1) + ... + T(n-2))  +  2 * T(n-1)
                                  
                                  # with
                                  T(n-1) = n-1 + 2 * (sum of T(x) for x=0 to n-2)
                                  T(n-1) = n-1 + 2 * (T(0) + T(1) + ... + T(n-2))
                                  
                                  # we can substitute:
                                  T(n) = 1 + T(n-1) + 2 * T(n-1)
                                  T(n) = 1 + 3 * T(n-1)
                                  
                                  S(0) = 1
                                  S(n) = 1 + S(n-1)
                                  
                                  T(0) = 1
                                  T(n) = O(n) + 2 * (sum of T(x) for x=0 to n-1)
                                  # and for simplicity, I will just use
                                  T(n) = n + 2 * (sum of T(x) for x=0 to n-1)
                                  
                                  T(1)  =   3 = 1 + 2 *  1  =  1 + 2 * (T(0))
                                  T(2)  =  10 = 2 + 2 *  4  =  2 + 2 * (T(0) + T(1))
                                  T(3)  =  31 = 3 + 2 * 14  =  3 + 2 * (T(0) + T(1) + T(2))
                                  T(4)  =  94 = 4 + 2 * 45  =  4 + 2 * (T(0) + T(1) + T(2) + T(3))
                                  
                                  T(n) = n + 2 * (sum of T(x) for x=0 to n-1)
                                  T(n) = n + 2 * (T(0) + T(1) + ... + T(n-1))
                                  T(n) = n + 2 * (T(0) + T(1) + ... + T(n-2)) + 2 * T(n-1)
                                  T(n) = 1 +  n-1 + 2 * (T(0) + T(1) + ... + T(n-2))  +  2 * T(n-1)
                                  
                                  # with
                                  T(n-1) = n-1 + 2 * (sum of T(x) for x=0 to n-2)
                                  T(n-1) = n-1 + 2 * (T(0) + T(1) + ... + T(n-2))
                                  
                                  # we can substitute:
                                  T(n) = 1 + T(n-1) + 2 * T(n-1)
                                  T(n) = 1 + 3 * T(n-1)
                                  
                                  S(0) = 1
                                  S(n) = 1 + S(n-1)
                                  

                                  Unable to resolve memory read and segmentation fault error in C++

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                                      void find_island(vector<vector<int>> & arg_grid, int* arg_coordinates) {
                                          if (arg_grid[*arg_coordinates][*(arg_coordinates + 1)] == INT_MAX) {
                                              *last = *arg_coordinates;
                                              *(last + 1) = *(arg_coordinates + 1);
                                              map_island(arg_grid, arg_coordinates, 1);
                                          } else {
                                              if (*(arg_coordinates + 1) < arg_grid[*arg_coordinates].size()) {
                                                  *(arg_coordinates + 1) = *(arg_coordinates + 1) + 1;
                                                  find_island(arg_grid, arg_coordinates);
                                              }
                                              if (*(arg_coordinates + 1) == arg_grid[*arg_coordinates].size()) {
                                                  if (*arg_coordinates == arg_grid.size()) return;
                                                  *arg_coordinates = *arg_coordinates + 1;
                                                  *(arg_coordinates + 1) = 0;
                                                  find_island(arg_grid, arg_coordinates);
                                              }
                                          }
                                      }
                                  
                                  *arg_coordinates = *arg_coordinates + 1;
                                  if (*arg_coordinates == arg_grid.size()) return;
                                  
                                      void find_island(vector<vector<int>> & arg_grid, int* arg_coordinates) {
                                          if (arg_grid[*arg_coordinates][*(arg_coordinates + 1)] == INT_MAX) {
                                              *last = *arg_coordinates;
                                              *(last + 1) = *(arg_coordinates + 1);
                                              map_island(arg_grid, arg_coordinates, 1);
                                          } else {
                                              if (*(arg_coordinates + 1) < arg_grid[*arg_coordinates].size()) {
                                                  *(arg_coordinates + 1) = *(arg_coordinates + 1) + 1;
                                                  find_island(arg_grid, arg_coordinates);
                                              }
                                              if (*(arg_coordinates + 1) == arg_grid[*arg_coordinates].size()) {
                                                  if (*arg_coordinates == arg_grid.size()) return;
                                                  *arg_coordinates = *arg_coordinates + 1;
                                                  *(arg_coordinates + 1) = 0;
                                                  find_island(arg_grid, arg_coordinates);
                                              }
                                          }
                                      }
                                  
                                  *arg_coordinates = *arg_coordinates + 1;
                                  if (*arg_coordinates == arg_grid.size()) return;
                                  

                                  Detecting signed integer multiplication overflow in C

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                                  reversed += (x % 10) * tenPower;
                                  
                                  ((x % 10) * tenPower) > INT_MAX
                                  
                                  (reversed + (x % 10) * tenPower) > INT_MAX
                                  
                                  if ((x % 10) != 0 && (tenPower > INT_MAX / (x % 10)) || 
                                                        (((x % 10) * tenPower) > INT_MAX - reversed))
                                      return 0;
                                  
                                  reversed += (x % 10) * tenPower;
                                  
                                  ((x % 10) * tenPower) > INT_MAX
                                  
                                  (reversed + (x % 10) * tenPower) > INT_MAX
                                  
                                  if ((x % 10) != 0 && (tenPower > INT_MAX / (x % 10)) || 
                                                        (((x % 10) * tenPower) > INT_MAX - reversed))
                                      return 0;
                                  
                                  reversed += (x % 10) * tenPower;
                                  
                                  ((x % 10) * tenPower) > INT_MAX
                                  
                                  (reversed + (x % 10) * tenPower) > INT_MAX
                                  
                                  if ((x % 10) != 0 && (tenPower > INT_MAX / (x % 10)) || 
                                                        (((x % 10) * tenPower) > INT_MAX - reversed))
                                      return 0;
                                  
                                  reversed += (x % 10) * tenPower;
                                  
                                  ((x % 10) * tenPower) > INT_MAX
                                  
                                  (reversed + (x % 10) * tenPower) > INT_MAX
                                  
                                  if ((x % 10) != 0 && (tenPower > INT_MAX / (x % 10)) || 
                                                        (((x % 10) * tenPower) > INT_MAX - reversed))
                                      return 0;
                                  
                                  int reverse(int n) {
                                      // 1st overflow checking...
                                      // Check if the absolute value is greater than INT32_MAX.
                                      // I converted "int" to "long" to get correct overflow value.
                                      if (n < 0 && (long) n * -1l > INT32_MAX) {
                                          return 0;
                                      }
                                      
                                      int res = 0;
                                      // Convert to absolute value.
                                      int tmp = n < 0 ? n * -1 : n;
                                      
                                      while (tmp > 0) {
                                          // Get the right most digit and add it to the current result.
                                          res += tmp % 10;
                                          // Remove the right most digit.
                                          tmp /= 10;
                                          
                                          // "tmp" still has remaining numbers.
                                          if (tmp > 0) {
                                              // 2nd overflow checking...
                                              // Check if reversed value will be greater than INT32_MAX when appending 0 to right most.
                                              // I converted "int" to "long" to get correct overflow value.
                                              if ((long) res * 10l > INT32_MAX) {
                                                  return 0;
                                              }
                                              
                                              // Append 0 to right most value of result.
                                              // If result is equal to 0, do not append 0.
                                              res *= res == 0 ? 1 : 10;
                                          }
                                      }
                                      
                                      // Return result.
                                      // If original value is negative, return negative result value..
                                      return n < 0 ? res * -1 : res;
                                  }
                                  
                                    if (x < 0) { pos = -x; }
                                    ...
                                    (x % 10) * tenPower
                                    ....
                                    reversed += (x % 10) * tenPower;
                                    ...
                                    if (x < 0) return -reversed;
                                  
                                    if (x < 0) { 
                                      if (x < -INT_MAX) { puts("Overflow"); return 0; }
                                      pos = -x;
                                    }
                                  
                                    // x is >= 0 here, tenPower >= 1
                                    int digit = x % 10;
                                    if (digit > INT_MAX/tenPower)     { puts("Overflow"); return 0; }
                                    int digit10 = digit * tenPower;
                                    if (reversed > INT_MAX - digit10) { puts("Overflow"); return 0; }
                                    reversed += digit10;
                                  
                                  // Return 1 on overflow
                                  int is_undefined_mult1(int a, int b) {
                                    if (a > 0) {
                                      if (b > 0) {
                                        return a > INT_MAX / b;       // a positive, b positive
                                      }
                                      return b < INT_MIN / a;         // a positive, b not positive
                                    }
                                    if (b > 0) {
                                      return a < INT_MIN / b;         // a not positive, b positive
                                    }
                                    return a != 0 && b < INT_MAX / a; // a not positive, b not positive
                                  }
                                  
                                    int digit = x % 10;
                                    if (is_undefined_mult1(digit, tenPower))  { puts("Overflow"); return 0; }
                                    int digit10 = digit * tenPower;
                                    if (is_undefined_add1(reversed, digit10)) { puts("Overflow"); return 0; }
                                    reversed += digit10;
                                  
                                    if (x < 0) { pos = -x; }
                                    ...
                                    (x % 10) * tenPower
                                    ....
                                    reversed += (x % 10) * tenPower;
                                    ...
                                    if (x < 0) return -reversed;
                                  
                                    if (x < 0) { 
                                      if (x < -INT_MAX) { puts("Overflow"); return 0; }
                                      pos = -x;
                                    }
                                  
                                    // x is >= 0 here, tenPower >= 1
                                    int digit = x % 10;
                                    if (digit > INT_MAX/tenPower)     { puts("Overflow"); return 0; }
                                    int digit10 = digit * tenPower;
                                    if (reversed > INT_MAX - digit10) { puts("Overflow"); return 0; }
                                    reversed += digit10;
                                  
                                  // Return 1 on overflow
                                  int is_undefined_mult1(int a, int b) {
                                    if (a > 0) {
                                      if (b > 0) {
                                        return a > INT_MAX / b;       // a positive, b positive
                                      }
                                      return b < INT_MIN / a;         // a positive, b not positive
                                    }
                                    if (b > 0) {
                                      return a < INT_MIN / b;         // a not positive, b positive
                                    }
                                    return a != 0 && b < INT_MAX / a; // a not positive, b not positive
                                  }
                                  
                                    int digit = x % 10;
                                    if (is_undefined_mult1(digit, tenPower))  { puts("Overflow"); return 0; }
                                    int digit10 = digit * tenPower;
                                    if (is_undefined_add1(reversed, digit10)) { puts("Overflow"); return 0; }
                                    reversed += digit10;
                                  
                                    if (x < 0) { pos = -x; }
                                    ...
                                    (x % 10) * tenPower
                                    ....
                                    reversed += (x % 10) * tenPower;
                                    ...
                                    if (x < 0) return -reversed;
                                  
                                    if (x < 0) { 
                                      if (x < -INT_MAX) { puts("Overflow"); return 0; }
                                      pos = -x;
                                    }
                                  
                                    // x is >= 0 here, tenPower >= 1
                                    int digit = x % 10;
                                    if (digit > INT_MAX/tenPower)     { puts("Overflow"); return 0; }
                                    int digit10 = digit * tenPower;
                                    if (reversed > INT_MAX - digit10) { puts("Overflow"); return 0; }
                                    reversed += digit10;
                                  
                                  // Return 1 on overflow
                                  int is_undefined_mult1(int a, int b) {
                                    if (a > 0) {
                                      if (b > 0) {
                                        return a > INT_MAX / b;       // a positive, b positive
                                      }
                                      return b < INT_MIN / a;         // a positive, b not positive
                                    }
                                    if (b > 0) {
                                      return a < INT_MIN / b;         // a not positive, b positive
                                    }
                                    return a != 0 && b < INT_MAX / a; // a not positive, b not positive
                                  }
                                  
                                    int digit = x % 10;
                                    if (is_undefined_mult1(digit, tenPower))  { puts("Overflow"); return 0; }
                                    int digit10 = digit * tenPower;
                                    if (is_undefined_add1(reversed, digit10)) { puts("Overflow"); return 0; }
                                    reversed += digit10;
                                  
                                  if digit.bit_length() >= 32:
                                          return 0
                                      else:
                                          return reversed
                                  

                                  Not being able to find an edge case in question &quot;Process Tasks Using Servers&quot; asked in leetcode weekly contest 243

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                                  serverPriority = [(name, index) for index, name in enumerate(servers)]
                                          
                                          serverPriority.sort()
                                  
                                          serverId = [i for i in range(len(servers))]
                                          heapq.heapify(serverId)
                                          
                                          res = [-1] * len(tasks)
                                          
                                          runningServers = []
                                          heapq.heapify(runningServers)
                                          
                                          currTime = 0
                                          j = 0
                                          while j < len(tasks):
                                              currTime = max(currTime, j)
                                              
                                              if len(runningServers) == 0:
                                                  ID = heapq.heappop(serverId)
                                                  res[j] = serverPriority[ID][1]
                                                  
                                                  heapq.heappush(runningServers, (currTime + tasks[j], ID))
                                              else:
                                                  # Here, we are first freeing up the servers
                                                  while len(runningServers) > 0 and currTime >= runningServers[0][0]:
                                                      time, ID = heapq.heappop(runningServers)
                                                      heapq.heappush(serverId, ID)
                                              
                                                  if len(serverId) > 0:
                                                      ID = heapq.heappop(serverId)
                                                      res[j] = serverPriority[ID][1]
                                  
                                                      heapq.heappush(runningServers, (currTime + tasks[j], ID))
                                                  else:
                                                      """
                                                       here, we are keeping record of the server 
                                                       with earliest finish time.
                                                      """
                                                      time = runningServers[0][0]
                                                      currTime = time
                                                      j -= 1
                                              
                                              j += 1
                                              
                                          return res
                                  

                                  Why does my operation of mod (%) on a 2 digit number return zero in python

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                                          if number >= 1000:
                                              thousands = number - (number % 1000)              # 3549 - (3549 % 1000) = 3000
                                  
                                              hundreds = number - thousands;                  #// 3549 - 3000 = 549
                                              tens = hundreds % 100                      #// 549 % 100 = 49
                                              hundreds = hundreds - (hundreds % 100)     #// 549 - (549 % 100) = 500
                                  
                                              unit = tens % 10 # you input "units" here
                                              tens = tens - (tens % 10)                  #// 49 - (49 % 10) = 49 - 9 = 9
                                  
                                              thou = thousands // 1000                     # 3
                                  
                                  MMMDXLIX
                                  
                                  def intToRoman(num: int) -> str:
                                      roman_dict_key = [1,4,5,9,10,40,50,90,100,400,500,900,1000][::-1]
                                      roman_dict_value = ["I","IV","V","IX","X","XL","L","XC","C","CD","D","CM","M"][::-1]
                                      roman_dict = dict(zip(roman_dict_key,roman_dict_value))
                                      ans = ""
                                      for digit,roman in roman_dict.items():
                                          while num>=digit:
                                              num -= digit
                                              ans += roman
                                      return ans
                                  
                                          if number >= 1000:
                                              thousands = number - (number % 1000)              # 3549 - (3549 % 1000) = 3000
                                  
                                              hundreds = number - thousands;                  #// 3549 - 3000 = 549
                                              tens = hundreds % 100                      #// 549 % 100 = 49
                                              hundreds = hundreds - (hundreds % 100)     #// 549 - (549 % 100) = 500
                                  
                                              unit = tens % 10 # you input "units" here
                                              tens = tens - (tens % 10)                  #// 49 - (49 % 10) = 49 - 9 = 9
                                  
                                              thou = thousands // 1000                     # 3
                                  
                                  MMMDXLIX
                                  
                                  def intToRoman(num: int) -> str:
                                      roman_dict_key = [1,4,5,9,10,40,50,90,100,400,500,900,1000][::-1]
                                      roman_dict_value = ["I","IV","V","IX","X","XL","L","XC","C","CD","D","CM","M"][::-1]
                                      roman_dict = dict(zip(roman_dict_key,roman_dict_value))
                                      ans = ""
                                      for digit,roman in roman_dict.items():
                                          while num>=digit:
                                              num -= digit
                                              ans += roman
                                      return ans
                                  
                                          if number >= 1000:
                                              thousands = number - (number % 1000)              # 3549 - (3549 % 1000) = 3000
                                  
                                              hundreds = number - thousands;                  #// 3549 - 3000 = 549
                                              tens = hundreds % 100                      #// 549 % 100 = 49
                                              hundreds = hundreds - (hundreds % 100)     #// 549 - (549 % 100) = 500
                                  
                                              unit = tens % 10 # you input "units" here
                                              tens = tens - (tens % 10)                  #// 49 - (49 % 10) = 49 - 9 = 9
                                  
                                              thou = thousands // 1000                     # 3
                                  
                                  MMMDXLIX
                                  
                                  def intToRoman(num: int) -> str:
                                      roman_dict_key = [1,4,5,9,10,40,50,90,100,400,500,900,1000][::-1]
                                      roman_dict_value = ["I","IV","V","IX","X","XL","L","XC","C","CD","D","CM","M"][::-1]
                                      roman_dict = dict(zip(roman_dict_key,roman_dict_value))
                                      ans = ""
                                      for digit,roman in roman_dict.items():
                                          while num>=digit:
                                              num -= digit
                                              ans += roman
                                      return ans
                                  

                                  What is wrong with my &quot;stars and bars&quot; implementation?

                                  copy iconCopydownload iconDownload
                                  '**|*|**|**'.scan(/(?<=\|)\*+(?=\|)/).sum(&:size) 
                                    #=> 3 
                                  
                                  def find_em(str, start_idx, end_idx)
                                     start_idx.zip(end_idx).map do |s,e|
                                       next 0 unless str[s..e].include?('|')
                                       str[str.index('|', s)..str.rindex('|', e)].count('*')
                                     end
                                  end
                                  
                                  find_em('**|*|**|**', [1, 1, 1, 0, 0], [6, 7, 8, 1, 3])
                                    #=> [1, 3, 3, 0, 0]
                                  
                                  '**|*|**|**'.scan(/(?<=\|)\*+(?=\|)/).sum(&:size) 
                                    #=> 3 
                                  
                                  def find_em(str, start_idx, end_idx)
                                     start_idx.zip(end_idx).map do |s,e|
                                       next 0 unless str[s..e].include?('|')
                                       str[str.index('|', s)..str.rindex('|', e)].count('*')
                                     end
                                  end
                                  
                                  find_em('**|*|**|**', [1, 1, 1, 0, 0], [6, 7, 8, 1, 3])
                                    #=> [1, 3, 3, 0, 0]
                                  
                                  '**|*|**|**'.scan(/(?<=\|)\*+(?=\|)/).sum(&:size) 
                                    #=> 3 
                                  
                                  def find_em(str, start_idx, end_idx)
                                     start_idx.zip(end_idx).map do |s,e|
                                       next 0 unless str[s..e].include?('|')
                                       str[str.index('|', s)..str.rindex('|', e)].count('*')
                                     end
                                  end
                                  
                                  find_em('**|*|**|**', [1, 1, 1, 0, 0], [6, 7, 8, 1, 3])
                                    #=> [1, 3, 3, 0, 0]
                                  

                                  Does Counter from collections maintain order after frequency has been considered, if all other frequencies are the same?

                                  copy iconCopydownload iconDownload
                                  >>> print(cnt.keys())
                                  dict_keys(['l', 'e', 't', 'c', 'o', 'd'])
                                  
                                  def __repr__(self):
                                      if not self:
                                          return f'{self.__class__.__name__}()'
                                      try:
                                          # dict() preserves the ordering returned by most_common()
                                          d = dict(self.most_common())
                                      except TypeError:
                                          # handle case where values are not orderable
                                          d = dict(self)
                                      return f'{self.__class__.__name__}({d!r})'
                                  
                                  >>> print(cnt.keys())
                                  dict_keys(['l', 'e', 't', 'c', 'o', 'd'])
                                  
                                  def __repr__(self):
                                      if not self:
                                          return f'{self.__class__.__name__}()'
                                      try:
                                          # dict() preserves the ordering returned by most_common()
                                          d = dict(self.most_common())
                                      except TypeError:
                                          # handle case where values are not orderable
                                          d = dict(self)
                                      return f'{self.__class__.__name__}({d!r})'
                                  

                                  Remove Nth Node From End of List LeetCode

                                  copy iconCopydownload iconDownload
                                  if not pointer1.next:
                                      return pointer1.next
                                  
                                  def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
                                      pointer1 = head
                                      pointer2 = head
                                      count = 0
                                      while pointer2 is not None:
                                          if count > n:
                                              pointer1 = pointer1.next
                                          pointer2 = pointer2.next
                                          count += 1
                                      if count > n:
                                          pointer1.next = pointer1.next.next
                                          return head  
                                      else:
                                          return head.next
                                  
                                  if not pointer1.next:
                                      return pointer1.next
                                  
                                  def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
                                      pointer1 = head
                                      pointer2 = head
                                      count = 0
                                      while pointer2 is not None:
                                          if count > n:
                                              pointer1 = pointer1.next
                                          pointer2 = pointer2.next
                                          count += 1
                                      if count > n:
                                          pointer1.next = pointer1.next.next
                                          return head  
                                      else:
                                          return head.next
                                  

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                                  QUESTION

                                  LeetCode TwoSum question returning buggy answer (C Implementation)

                                  Asked 2021-Jun-15 at 16:06

                                  Here is the question:

                                  Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.
                                  
                                  You may assume that each input would have exactly one solution, and you may not use the same element twice.
                                  
                                  You can return the answer in any order
                                  

                                  Here is my solution for the same:

                                  int* twoSum(int* nums, int numsSize, int target, int* returnSize){
                                      
                                      int* result= (int*)malloc(2*sizeof(int));
                                      int* iterator= nums;
                                      int sum=0;
                                      
                                      for (int i=0; i<numsSize; i++)
                                      {
                                          for (int j=0; j<numsSize; j++)
                                          {
                                              if (i!=j)
                                              { 
                                                  sum=iterator[i]+iterator[j];
                                                  if (sum==target)
                                                  {
                                                      result[0]=i;
                                                      result[1]=j;
                                                      return result;
                                                  }
                                              }
                                          }
                                      }
                                  
                                      return NULL;
                                  
                                  }
                                  
                                  

                                  I know that it is not the most optimal solution, but I assumed it would work. LeetCode shows me that the answer returned is [] or ]. It seems to work fine on VSCode. I was wondering if I could get some help regarding what the error might be.

                                  ANSWER

                                  Answered 2021-Jun-15 at 16:06

                                  You have to assign the number of elements of the array to return (2 in this case) to what the argument returnSize points at (*returnSize) to tell the judge system how large the array you returned is.

                                  Source https://stackoverflow.com/questions/67989752

                                  Community Discussions, Code Snippets contain sources that include Stack Exchange Network

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