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Hello **Programmers/Coders,** Today we are going to share ** solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python**. At Each Problem with Successful submission with

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In this post, you will find the solution for the **Unique Paths** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** –

**Unique Paths**– LeetCode Problem

Unique Paths– LeetCode Problem

**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 * 10`

.^{9}

**Example 1:**

Input:m = 3, n = 7Output:28

**Example 2:**

Input:m = 3, n = 2Output:3Explanation: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:**

Unique Paths– LeetCode Solutions

Unique Paths in C++:

class Solution { public: int uniquePaths(int m, int n) { vectordp(n, 1); for (int i = 1; i < m; ++i) for (int j = 1; j < n; ++j) dp[j] += dp[j - 1]; return dp[n - 1]; } };

Unique Paths in Java:

class Solution { public int uniquePaths(int m, int n) { int[] dp = new int[n]; Arrays.fill(dp, 1); for (int i = 1; i < m; ++i) for (int j = 1; j < n; ++j) dp[j] += dp[j - 1]; return dp[n - 1]; } }

Unique Paths 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]