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LeetCode-Solutions | Weekly Update ) Python / Modern C Solutions of All | Learning library

 by   kamyu104 C++ Version: Current License: MIT

 by   kamyu104 C++ Version: Current License: MIT

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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
Quality
Security
Security
License
License
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kandi-support Support

  • LeetCode-Solutions has a medium active ecosystem.
  • It has 2519 star(s) with 884 fork(s). There are 120 watchers for this library.
  • It had no major release in the last 12 months.
  • There are 0 open issues and 33 have been closed. On average issues are closed in 0 days. There are 24 open pull requests and 0 closed requests.
  • It has a neutral sentiment in the developer community.
  • The latest version of LeetCode-Solutions is current.
LeetCode-Solutions Support
Best in #Learning
Average in #Learning
LeetCode-Solutions Support
Best in #Learning
Average in #Learning

quality kandi Quality

  • LeetCode-Solutions has 0 bugs and 0 code smells.
LeetCode-Solutions Quality
Best in #Learning
Average in #Learning
LeetCode-Solutions Quality
Best in #Learning
Average in #Learning

securitySecurity

  • 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.
LeetCode-Solutions Security
Best in #Learning
Average in #Learning
LeetCode-Solutions Security
Best in #Learning
Average in #Learning

license 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.
LeetCode-Solutions License
Best in #Learning
Average in #Learning
LeetCode-Solutions License
Best in #Learning
Average in #Learning

buildReuse

  • LeetCode-Solutions releases are not available. You will need to build from source code and install.
LeetCode-Solutions Reuse
Best in #Learning
Average in #Learning
LeetCode-Solutions Reuse
Best in #Learning
Average in #Learning
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LeetCode-Solutions Key Features

R.I.P. to my old Leetcode repository, where there were 5.7k+ stars and 2.2k+ forks (ever the top 3 in the field).

Since free questions may be even mistakenly taken down by some companies, only solutions will be post on now.

There are new LeetCode questions every week. I'll keep updating for full summary and better solutions.

For more problem solutions, you can see my LintCode, GoogleKickStart repositories.

For more challenging problem solutions, you can also see my GoogleCodeJam, FacebookHackerCup repositories.

Hope you enjoy the journey of learning data structures and algorithms.

Notes: "🔒" means your subscription of LeetCode premium membership is required for reading the question.

Community Discussions

Trending Discussions on LeetCode-Solutions
  • Why this backtracking solution for Unique Path problem is O(2^max(m,n))?
Trending Discussions on LeetCode-Solutions

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

import java.util.*;
import java.lang.*;
import java.io.*;
class Solution {
   public static int uniquePaths(int m, int n) {
       if(m == 1 || n == 1)
           return 1;
       return uniquePaths(m-1, n) + uniquePaths(m, n-1);
   }
   
   public static void main(String[] args){
     System.out.print(uniquePaths(2,2));
   }
}

The worst time complexity is mentioned in the wbesite as O(2^max(m,n)). It looks incorrect to me.

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

T(mn) = T(mn-1) + T(mn-1)
     = 2 * T(mn-1)
     = 2^mn

So worst time complexity would be O(2^mn). Let me know if my calculation is correct or if I am missing something

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:

T(mn) = T(m(n-1)) + T((m-1)n) + 1

This reflects your recursive calls: the product of the two arguments (that are passed to the recursive call) represents the size of the remaining problem space.

For calculating the time complexity it is important to add the constant term, as each call of the function represents work/time.

Furthermore, with this notation the distinction is gone between products that are the same, but come from different grid sizes. You should keep the 2 dimensions separate:

T(m, n) = T(m, n-1) + T(m-1, n) + 1
T(m, 1) = 1
T(1, n) = 1

The base cases represent the case where only 1 column or 1 row is left, and then the robot can only move in one direction -- via one path.

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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