LeetCode | : pencil2 : LeetCode solutions in C++ 11 and Python3 | Natural Language Processing library

 by   pezy C++ Version: Current License: MIT

kandi X-RAY | LeetCode Summary

kandi X-RAY | LeetCode Summary

LeetCode is a C++ library typically used in Artificial Intelligence, Natural Language Processing, LeetCode applications. LeetCode has no bugs, it has no vulnerabilities, it has a Permissive License and it has medium support. You can download it from GitHub.

leetcode solutions in c++ 11 and python3. |no.|title|solution|note|difficulty|tag| |---|-----|--------|----|----------|---| |0|[two sum] c(000.%20two%20sum/solution.h) [python] 000.%20two%20sum/solution.py)|[note] 000.%20two%20sum)|easy|mapping| |1|[add two numbers] c(001.%20add%20two%20numbers/solution.h) [python] 001.%20add%20two%20numbers/solution.py)|[note] 001.%20add%20two%20numbers)|medium|linkedlist| |2|[longest substring without repeating characters] c(002.%20longest%20substring%20without%20repeating%20characters/solution.h) [python] 002.%20longest%20substring%20without%20repeating%20characters/solution.py)|[note] 002.%20longest%20substring%20without%20repeating%20characters)|medium|mapping| |3|[median of two sorted arrays] c(003.%20median%20of%20two%20sorted%20arrays/solution.h) [python] 003.%20median%20of%20two%20sorted%20arrays/solution.py)|[note] 003.%20median%20of%20two%20sorted%20arrays)|hard| |4|[longest palindromic substring] c(004.%20longest%20palindromic%20substring/solution.h) [python] 004.%20longest%20palindromic%20substring/solution.py)|[note] 004.%20longest%20palindromic%20substring)|medium| |5|[zigzag conversion] c(005.%20zigzag%20conversion/solution.h) [python] 005.%20zigzag%20conversion/solution.py)|[note] 005.%20zigzag%20conversion)|medium| |6|[reverse integer] c(006.%20reverse%20integer/solution.h) [python] 006.%20reverse%20integer/solution.py)|[note] 006.%20reverse%20integer)|easy| |7|[string to integer (atoi)] c(007.%20string%20to%20integer%20(atoi)/solution.h) [python] 007.%20string%20to%20integer%20(atoi)/solution.py)|[note] 007.%20string%20to%20integer%20(atoi))|medium| |8|[palindrome number] c(008.%20palindrome%20number/solution.h) [python] 008.%20palindrome%20number/solution.py)|[note] 008.%20palindrome%20number)|easy| |9|[regular expression matching] c(009.%20regular%20expression%20matching/solution.h) [python] 009.%20regular%20expression%20matching/solution.py)|[note] 009.%20regular%20expression%20matching)|hard| |10|[container with most water] c(010.%20container%20with%20most%20water/solution.h) [python] 010.%20container%20with%20most%20water/solution.py)|[note] 010.%20container%20with%20most%20water)|medium| |11|[integer to roman] c(011.%20integer%20to%20roman/solution.h) [python] 011.%20integer%20to%20roman/solution.py)|[note] 011.%20integer%20to%20roman)|medium| |12|[roman to integer] c(012.%20roman%20to%20integer/solution.h) [python] 012.%20roman%20to%20integer/solution.py)|[note] 012.%20roman%20to%20integer)|easy| |13|[longest common prefix] c(013.%20longest%20common%20prefix/solution.h) [python] 013.%20longest%20common%20prefix/solution.py)|[note] 013.%20longest%20common%20prefix)|easy| |14|[3sum] c(014.%203sum/solution.h) [python] 014.%203sum/solution.py)|[note] 014.%203sum)|medium| |15|[3sum closest] c(015.%203sum%20closest/solution.h) [python] 015.%203sum%20closest/solution.py)|[note] 015.%203sum%20closest)|medium| |16|[letter combinations of a phone number] c(016.%20letter%20combinations%20of%20a%20phone%20number/solution.h) [python] 016.%20letter%20combinations%20of%20a%20phone%20number/solution.py)|[note] 016.%20letter%20combinations%20of%20a%20phone%20number)|medium| |17|[4sum] c(017.%204sum/solution.h) [python] 017.%204sum/solution.py)|[note] 017.%204sum)|medium| |18|[remove nth node from end of list] c(018.%20remove%20nth%20node%20from%20end%20of%20list/solution.h) [python] 018.%20remove%20nth%20node%20from%20end%20of%20list/solution.py)|[note] 018.%20remove%20nth%20node%20from%20end%20of%20list)|medium| |19|[valid parentheses] c(019.%20valid%20parentheses/solution.h) [python] 019.%20valid%20parentheses/solution.py)|[note] 019.%20valid%20parentheses)|easy| |20|[merge two sorted lists] c(020.%20merge%20two%20sorted%20lists/solution.h) [python] 020.%20merge%20two%20sorted%20lists/solution.py)|[note] 020.%20merge%20two%20sorted%20lists)|easy| |21|[generate parentheses] c(021.%20generate%20parentheses/solution.h) [python] 021.%20generate%20parentheses/solution.py)|[note] 021.%20generate%20parentheses)|medium| |22|[merge k sorted lists] c(022.%20merge%20k%20sorted%20lists/solution.h) [python] 022.%20merge%20k%20sorted%20lists/solution.py)|[note] 022.%20merge%20k%20sorted%20lists)|hard| |23|[swap nodes in pairs] c(023.%20swap%20nodes%20in%20pairs/solution.h) [python] 023.%20swap%20nodes%20in%20pairs/solution.py)|[note] 023.%20swap%20nodes%20in%20pairs)|medium| |24|[reverse nodes in k-group] c(024.%20reverse%20nodes%20in%20k-group/solution.h) [python] 024.%20reverse%20nodes%20in%20k-group/solution.py)|[note] 024.%20reverse%20nodes%20in%20k-group)|hard| |25|[remove duplicates from sorted array] c(025.%20remove%20duplicates%20from%20sorted%20array/solution.h) [python] 025.%20remove%20duplicates%20from%20sorted%20array/solution.py)|[note] 025.%20remove%20duplicates%20from%20sorted%20array)|easy| |26|[remove element] c(026.%20remove%20element/solution.h) [python] 026.%20remove%20element/solution.py)|[note] 026.%20remove%20element)|easy| |27|[implement strstr()] c(027.%20implement%20strstr()/solution.h) [python] 027.%20implement%20strstr()/solution.py)|[note] 027.%20implement%20strstr())|easy| |28|[divide two integers] c(028.%20divide%20two%20integers/solution.h) [python] 028.%20divide%20two%20integers/solution.py)|[note] 028.%20divide%20two%20integers)|medium| |29|[substring with concatenation of all words] c(029.%20substring%20with%20concatenation%20of%20all%20words/solution.h) [python] 029.%20substring%20with%20concatenation%20of%20all%20words/solution.py)|[note] 029.%20substring%20with%20concatenation%20of%20all%20words)|hard| |30|[next permutation] c(030.%20next%20permutation/solution.h) [python] 030.%20next%20permutation/solution.py)|[note] 030.%20next%20permutation)|medium| |31|[longest valid parentheses] c(031.%20longest%20valid%20parentheses/solution.h) [python] 031.%20longest%20valid%20parentheses/solution.py)|[note] 031.%20longest%20valid%20parentheses)|hard| |32|[search in rotated sorted array] c(032.%20search%20in%20rotated%20sorted%20array/solution.h) [python] 032.%20search%20in%20rotated%20sorted%20array/solution.py)|[note] 032.%20search%20in%20rotated%20sorted%20array)|medium| |33|[search for a range] c(033.%20search%20for%20a%20range/solution.h) [python] 033.%20search%20for%20a%20range/solution.py)|[note] 033.%20search%20for%20a%20range)|medium| |34|[search insert position] c(034.%20search%20insert%20position/solution.h) [python] 034.%20search%20insert%20position/solution.py)|[note] 034.%20search%20insert%20position)|easy| |35|[valid sudoku] c(035.%20valid%20sudoku/solution.h) [python] 035.%20valid%20sudoku/solution.py)|[note] 035.%20valid%20sudoku)|medium| |36|[sudoku solver] c(036.%20sudoku%20solver/solution.h) [python] 036.%20sudoku%20solver/solution.py)|[note] 036.%20sudoku%20solver)|hard| |37|[count and say] c(037.%20count%20and%20say/solution.h) [python] 037.%20count%20and%20say/solution.py)|[note] 037.%20count%20and%20say)|easy| |38|[combination sum] c(038.%20combination%20sum/solution.h) [python] 038.%20combination%20sum/solution.py)|[note] 038.%20combination%20sum)|medium| |39|[combination sum ii] c(039.%20combination%20sum%20ii/solution.h) [python] 039.%20combination%20sum%20ii/solution.py)|[note] 039.%20combination%20sum%20ii)|medium| |40|[first missing positive] c(040.%20first%20missing%20positive/solution.h) [python] 040.%20first%20missing%20positive/solution.py)|[note] 040.%20first%20missing%20positive)|hard| |41|[trapping rain water] c(041.%20trapping%20rain%20water/solution.h) [python] 041.%20trapping%20rain%20water/solution.py)|[note] 041.%20trapping%20rain%20water)|hard| |42|[multiply strings] c(042.%20multiply%20strings/solution.h) [python] 042.%20multiply%20strings/solution.py)|[note] 042.%20multiply%20strings)|medium| |43|[wildcard matching] c(043.%20wildcard%20matching/solution.h) [python] 043.%20wildcard%20matching/solution.py)|[note] 043.%20wildcard%20matching)|hard| |44|[jump game ii] c(044.%20jump%20game%20ii/solution.h) [python] 044.%20jump%20game%20ii/solution.py)|[note] 044.%20jump%20game%20ii)|hard| |45|[permutations] c(045.%20permutations/solution.h) [python] 045.%20permutations/solution.py)|[note] 045.%20permutations)|medium| |46|[permutations ii] c(046.%20permutations%20ii/solution.h) [python] 046.%20permutations%20ii/solution.py)|[note] 046.%20permutations%20ii)|medium| |47|[rotate

            kandi-support Support

              LeetCode has a medium active ecosystem.
              It has 3144 star(s) with 966 fork(s). There are 240 watchers for this library.
              It had no major release in the last 6 months.
              There are 2 open issues and 17 have been closed. On average issues are closed in 59 days. There are 3 open pull requests and 0 closed requests.
              It has a neutral sentiment in the developer community.
              The latest version of LeetCode is current.

            kandi-Quality Quality

              LeetCode has 0 bugs and 0 code smells.

            kandi-Security Security

              LeetCode has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.
              LeetCode code analysis shows 0 unresolved vulnerabilities.
              There are 0 security hotspots that need review.

            kandi-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.

            kandi-Reuse Reuse

              LeetCode releases are not available. You will need to build from source code and install.

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            LeetCode Key Features

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            LeetCode Examples and Code Snippets

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            Community Discussions


            Using while let with two variables simultaneously
            Asked 2022-Apr-10 at 08:55

            I'm learning Rust and have been going through leetcode problems. One of them includes merging two linked lists, whose nodes are optional. I want to write a while loop that would go on until at least 1 node becomes None, and I was trying to use the while let loop for that.

            However, it looks like the while let syntax supports only one optional, e.g.:



            Answered 2022-Apr-10 at 08:22


            F2 rename variable doesn't work in vscode + jupyter notebook + python
            Asked 2022-Mar-23 at 04:56

            I can use the normal F2 rename variable functionality in regular python files in vscode. But not when editing python in a jupyter notebook.

            When I press F2 on a variable in a jupyter notebook in vscode I get the familiar change variable window but when I press enter the variable is not changed and I get this error message:

            No result. No result.

            Is there a way to get the F2 change variable functionality to work in jupyter notebooks?

            Here's my system info:

            jupyter module version



            Answered 2022-Jan-17 at 02:49

            Notice that you put up a bug report in GitHub and see this issue: Renaming variables didn't work, the programmer replied:

            Some language features are currently not supported in notebooks, but we are making plans now to hopefully bring more of those online soon.

            So please wait for this feature.

            Source https://stackoverflow.com/questions/70729502


            Is this Union Find really O(n) as they claim?
            Asked 2022-Mar-14 at 07:33

            I am solving a problem on LeetCode:

            Given an unsorted array of integers nums, return the length of the longest consecutive elements sequence. You must write an algorithm that runs in O(n) time. So for nums = [100,4,200,1,3,2], the output is 4.

            The Union Find solution to solve this is as below:



            Answered 2022-Mar-14 at 07:33

            They are right. A properly implemented Union Find with path compression and union by rank has linear run time complexity as a whole, while any individual operation has an amortized constant run time complexity. The exact complexity of m operations of any type is O(m * alpha(n)) where alpha is the inverse Ackerman function. For any possible n in the physical world, the inverse Ackerman function doesn't exceed 4. Thus, we can state that individual operations are constant and algorithm as a whole linear.

            The key part for path compression in your code is here:

            Source https://stackoverflow.com/questions/71453109


            Are java streams able to lazilly reduce from map/filter conditions?
            Asked 2022-Jan-12 at 09:30

            I am using a functional programming style to solve the Leetcode easy question, Count the Number of Consistent Strings. The premise of this question is simple: count the amount of values for which the predicate of "all values are in another set" holds.

            I have two approaches, one which I am fairly certain behaves as I want it to, and the other which I am less sure about. Both produce the correct output, but ideally they would stop evaluating other elements after the output is in a final state.



            Answered 2022-Jan-12 at 09:30

            The actual term you’re asking for is short-circuiting

            Further, some operations are deemed short-circuiting operations. An intermediate operation is short-circuiting if, when presented with infinite input, it may produce a finite stream as a result. A terminal operation is short-circuiting if, when presented with infinite input, it may terminate in finite time. Having a short-circuiting operation in the pipeline is a necessary, but not sufficient, condition for the processing of an infinite stream to terminate normally in finite time.

            The term “lazy” only applies to intermediate operations and means that they only perform work when being requested by the terminal operation. This is always the case, so when you don’t chain a terminal operation, no intermediate operation will ever process any element.

            Finding out whether a terminal operation is short-circuiting, is rather easy. Go to the Stream API documentation and check whether the particular terminal operation’s documentation contains the sentence

            This is a short-circuiting terminal operation.

            allMatch has it, reduce has not.

            This does not mean that such optimizations based on logic or algebra are impossible. But the responsibility lies at the JVM’s optimizer which might do the same for loops. However, this requires inlining of all involved methods to be sure that this conditions always applies and there are no side effect which must be retained. This behavioral compatibility implies that even if the processing gets optimized away, a peek(System.out::println) would keep printing all elements as if they were processed. In practice, you should not expect such optimizations, as the Stream implementation code is too complex for the optimizer.

            Source https://stackoverflow.com/questions/70673795


            How are javascript variables referenced?
            Asked 2021-Dec-19 at 00:30

            I'm currently studying for a technical interview and I'm just going through the leetcode grind. I came across a question that is apparently asked pretty frequently by the company I'm about to interview at so I attempted it. I couldn't quite get it so I looked to the solution and came across this solution.



            Answered 2021-Dec-19 at 00:30

            This happens because prev's reference is being pointed by res when you do res=[prev], basically the address where the actual prev array is stored is pointed, as the prev updates, it also show changes in res.

            Source https://stackoverflow.com/questions/70408122


            Will std::sort always compare equal values?
            Asked 2021-Dec-18 at 21:29

            I am doing the following problem on leetcode: https://leetcode.com/problems/contains-duplicate/

            Given an integer array nums, return true if any value appears at least twice in the array, and return false if every element is distinct.

            The solution I came up to the problem is the following:



            Answered 2021-Dec-18 at 21:29

            Will std::sort always compare equal values or sometimes it can skip comparing them and therefore duplicate values will not be found?

            Yes, some equal value elements will always be compared if duplicates do exist.

            Let us assume the opposite: initial array of elements {e} for sorting contains a subset of elements having the same value and a valid sorting algorithm does not call comparison operator < for any pair of the elements from the subset.

            Then we construct same sized array of tuples {e,k}, with the first tuple value from the initial array and arbitrary selected second tuple value k, and apply the same sorting algorithm using the lexicographic comparison operator for the tuples. The order of tuples after sorting can deviate from the order of sorted elements {e} only for same value elements, where in the case of array of tuples it will depend on second tuple value k.

            Since we assumed that the sorting algorithm does not compare any pair of same value elements, then it will not compare the tuples with the same first tuple value, so the algorithm will be unable to sort them properly. This contradicts our assumptions and proves that some equal value elements (if they exist in the array) will always be compared during sorting.

            Source https://stackoverflow.com/questions/70136586


            How can I label connected components in APL?
            Asked 2021-Dec-16 at 12:26

            I'm trying to do leet puzzle https://leetcode.com/problems/max-area-of-island/, requiring labelling connected (by sides, not corners) components.

            How can I transform something like



            Answered 2021-Dec-16 at 12:26

            We can start off by enumerating the ones. We do the by applying the function (where, but since all are 1s, it is equivalent to 1,2,3,…) @ at the subset masked by the bits themselves, i.e. ⍸@⊢:

            Source https://stackoverflow.com/questions/70348704


            Understanding left hand notation of C(n,2)= n(n−1)​ / 2
            Asked 2021-Dec-13 at 16:22

            For an array of n integers, there are C(n,2)= n(n−1)​ / 2 pairs of integers. Thus, we may check all n(n−1)​ / 2 pairs and see if there is any pair with duplicates.

            I was poking around a LeetCode question and the answer for one of the algorithms included the above formula in the question explanation.

            What is the point of the C(n, 2) nomenclature on the left hand side of the equation? Is this a known/named standard that I can read and interpret, or is this some more general information that must/should be ascertained from context? I understand the math on the right, but I don't have any preconceived notions that adds any detail to my understanding from the function on the left.

            What is the 2 doing?



            Answered 2021-Dec-13 at 06:30

            It's called binomial coefficient, or "nCk" or "n Choose k".

            The formula is

            Here n is the size of the set, and k = 2 is the number of elements to select, so that e.g. sets {3, 6} and {6,3} taken are considered equal.

            AFAIK, the standard notation in combinatorics is as shown above and spelled "n choose k", where as C(...) is non-standard requiring clarification when first introduced.

            Source https://stackoverflow.com/questions/70329969


            Longest Common Prefix in Javascript
            Asked 2021-Dec-04 at 10:26

            I am trying to solve the Leet Code challenge 14. Longest Common Prefix:

            Write a function to find the longest common prefix string amongst an array of strings.

            If there is no common prefix, return an empty string "".

            Example 1: ...


            Answered 2021-Aug-09 at 00:42


            Binary Search in JS: trying to find a consistent mental model
            Asked 2021-Oct-11 at 13:28

            I am grinding LeetCode these days and I encountered the challenge 162. Find Peak Element:

            A peak element is an element that is strictly greater than its neighbors.

            Given an integer array nums, find a peak element, and return its index. If the array contains multiple peaks, return the index to any of the peaks.

            You may imagine that nums[-1] = nums[n] = -∞.

            You must write an algorithm that runs in O(log n) time.

            • 1 <= nums.length <= 1000
            • -231 <= nums[i] <= 231 - 1
            • nums[i] != nums[i + 1] for all valid i

            This question is about using binary search to find a peak element in an array.

            I know we can think of the array as alternating ascending and descending sequences. Here is my solution



            Answered 2021-Oct-11 at 06:20

            When the following condition is true:

            Source https://stackoverflow.com/questions/69456782

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