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In this post, you will find the solution for the **Minimum Path Sum** **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** – Minimum Path Sum– LeetCode Problem

Minimum Path Sum– LeetCode Problem

**Problem:**

Given a `m x n`

`grid`

filled with non-negative numbers, find a path from top left to bottom right, which minimizes the sum of all numbers along its path.

**Note:** You can only move either down or right at any point in time.

**Example 1:**

Input:grid = [[1,3,1],[1,5,1],[4,2,1]]Output:7Explanation:Because the path 1 → 3 → 1 → 1 → 1 minimizes the sum.

**Example 2:**

Input:grid = [[1,2,3],[4,5,6]]Output:12

**Constraints:**

`m == grid.length`

`n == grid[i].length`

`1 <= m, n <= 200`

`0 <= grid[i][j] <= 100`

Minimum Path Sum– LeetCode Solutions

Minimum Path Sum in C++:

class Solution { public: int minPathSum(vector>& grid) { const int m = grid.size(); const int n = grid[0].size(); for (int i = 0; i < m; ++i) for (int j = 0; j < n; ++j) if (i > 0 && j > 0) grid[i][j] += min(grid[i - 1][j], grid[i][j - 1]); else if (i > 0) grid[i][0] += grid[i - 1][0]; else if (j > 0) grid[0][j] += grid[0][j - 1]; return grid[m - 1][n - 1]; } };

Minimum Path Sum in Java:

class Solution { public int minPathSum(int[][] grid) { final int m = grid.length; final int n = grid[0].length; for (int i = 0; i < m; ++i) for (int j = 0; j < n; ++j) if (i > 0 && j > 0) grid[i][j] += Math.min(grid[i - 1][j], grid[i][j - 1]); else if (i > 0) grid[i][0] += grid[i - 1][0]; else if (j > 0) grid[0][j] += grid[0][j - 1]; return grid[m - 1][n - 1]; } }

Minimum Path Sum in Python:

class Solution: def minPathSum(self, grid: List[List[int]]) -> int: m = len(grid) n = len(grid[0]) for i in range(m): for j in range(n): if i > 0 and j > 0: grid[i][j] += min(grid[i - 1][j], grid[i][j - 1]) elif i > 0: grid[i][0] += grid[i - 1][0] elif j > 0: grid[0][j] += grid[0][j - 1] return grid[m - 1][n - 1]