LeetCode-Solutions | 🏋️ Python / Modern C++ Solutions of All 2719 LeetCode | Learning library

by kamyu104 C++ Version: Current License: MIT

X-Ray Key Features Code Snippets(3)Community Discussions(1)Vulnerabilities Install Support

kandi X-RAY | LeetCode-Solutions Summary

LeetCode-Solutions is a C++ library typically used in Tutorial, Learning, Example Codes, LeetCode applications. LeetCode-Solutions has no bugs, it has no vulnerabilities, it has a Permissive License and it has medium support. You can download it from GitHub.

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Support

Quality

Security

License

Reuse

Support

LeetCode-Solutions has a medium active ecosystem.

It has 4069 star(s) with 1418 fork(s). There are 166 watchers for this library.

It had no major release in the last 6 months.

There are 0 open issues and 46 have been closed. On average issues are closed in 0 days. There are 40 open pull requests and 0 closed requests.

It has a neutral sentiment in the developer community.

The latest version of LeetCode-Solutions is current.

Quality

LeetCode-Solutions has 0 bugs and 0 code smells.

Security

LeetCode-Solutions has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.

LeetCode-Solutions code analysis shows 0 unresolved vulnerabilities.

There are 0 security hotspots that need review.

License

LeetCode-Solutions is licensed under the MIT License. This license is Permissive.

Permissive licenses have the least restrictions, and you can use them in most projects.

Reuse

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

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LeetCode-Solutions Key Features

No Key Features are available at this moment for LeetCode-Solutions.

LeetCode-Solutions Examples and Code Snippets

Evaluates the r p and returns the results .

javascript

Lines of Code : 19

License : No License

Copy

function evalRPN(tokens) {
  let stack = [];

  for (let n of tokens) {
    // if operator function exists, 
    // execute it on the two most recent numbers
    if (map[n]) {
      let fn = map[n];
      let y = stack.pop();
      let x = stack.pop(

Calculates the next t in a loop

javascript

Lines of Code : 10

License : No License

Copy

function f(x) {
        if (x === null)
            return 0;
        var t = x.next;
        var tn = f(t);
        if (tn === n)
            x.next = t.next;

        return tn + 1;
    }

Generate a key .

javascript

Lines of Code : 8

License : No License

Copy

function generateKey (str) {
    let result = 1;
    for (let i = 0, l = str.length; i < l; i++) {
        let n = str[i].charCodeAt() - 'a'.charCodeAt() + 1;
        result = result * (n*n + n + 41) % 2147483647;
    }
    return result;
}

Community Discussions

Trending Discussions on LeetCode-Solutions

Why this backtracking solution for Unique Path problem is O(2^max(m,n))?

QUESTION

Why this backtracking solution for Unique Path problem is O(2^max(m,n))?

Asked 2022-Jan-14 at 17:15

This is unique path problem from LeetCode https://leetcode.com/problems/unique-paths/.

There is a robot on an m x n grid. The robot is initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]). The robot can only move either down or right at any point in time.

Given the two integers m and n, return the number of possible unique paths that the robot can take to reach the bottom-right corner.

Here is the backtracking solution from tutorialcup https://www.tutorialcup.com/leetcode-solutions/unique-paths-leetcode-solution.htm

...

ANSWER

Answered 2022-Jan-14 at 17:15

I think there is m*n possibilities which is reduced by one in each recursive step.

No it is reduced by either n or m in each step. So:

Source https://stackoverflow.com/questions/70713780

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

Vulnerabilities

No vulnerabilities reported

Install LeetCode-Solutions

You can download it from GitHub.

Support

For any new features, suggestions and bugs create an issue on GitHub. If you have any questions check and ask questions on community page Stack Overflow .

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