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

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: 7 Explanation: 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
Now, let’s see the leetcode solution of Minimum Path Sum Leetcode Solution.
Minimum Path Sum Leetcode Solution 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]
Minimum Path Sum Leetcode Solution in CPP
class Solution { public: int minPathSum(vector<vector<int>>& 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 Leetcode Solution 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]; } }
Note: This problem Minimum Path Sum is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.