In this post, we are going to solve the Unique Paths Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

Problem
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.
The test cases are generated so that the answer will be less than or equal to 2 * 109
.
Example 1:
Input: m = 3, n = 7 Output: 28
Example 2:
Input: m = 3, n = 2 Output: 3 Explanation: From the top-left corner, there are a total of 3 ways to reach the bottom-right corner: 1. Right -> Down -> Down 2. Down -> Down -> Right 3. Down -> Right -> Down
Constraints:
1 <= m, n <= 100
Now, let’s see the leetcode solution of Unique Paths Leetcode Solution.
Unique Paths Leetcode Solution in Python
class Solution: def uniquePaths(self, m: int, n: int) -> int: # dp[i][j] := unique paths from (0, 0) to (i, j) dp = [[1] * n for _ in range(m)] for i in range(1, m): for j in range(1, n): dp[i][j] = dp[i - 1][j] + dp[i][j - 1] return dp[-1][-1]
Unique Paths Leetcode Solution in CPP
class Solution { public: int uniquePaths(int m, int n) { // dp[i][j] := unique paths from (0, 0) to (i, j) vector<vector<int>> dp(m, vector<int>(n, 1)); for (int i = 1; i < m; ++i) for (int j = 1; j < n; ++j) dp[i][j] = dp[i - 1][j] + dp[i][j - 1]; return dp[m - 1][n - 1]; } };
Unique Paths Leetcode Solution in Java
class Solution { public int uniquePaths(int m, int n) { // dp[i][j] := unique paths from (0, 0) to (i, j) int[][] dp = new int[m][n]; Arrays.stream(dp).forEach(row -> Arrays.fill(row, 1)); for (int i = 1; i < m; ++i) for (int j = 1; j < n; ++j) dp[i][j] = dp[i - 1][j] + dp[i][j - 1]; return dp[m - 1][n - 1]; } }
Note: This problem Unique Paths is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.