Leetcode Minimum Path Sum problem solution YASH PAL, 31 July 2024 In this Leetcode Minimum Path Sum problem solution we have 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. Topics we are covering Toggle Problem solution in Python.Problem solution in Java.Problem solution in C++.Problem solution in C. Problem solution in Python. class Solution: def minPathSum(self, a: List[List[int]]) -> int: if len(a)==0: return 0 m=len(a) n=len(a[0]) c=[[0 for i in range(n)]for j in range(m)] c[0][0] = a[0][0] for i in range(1, n): c[0][i] = a[0][i]+c[0][i-1] for i in range(1, m): c[i][0] = a[i][0]+c[i-1][0] for i in range(1,m): for j in range(1,n): c[i][j] = min(c[i-1][j], c[i][j-1]) + a[i][j] return c[m-1][n-1] Problem solution in Java. import java.lang.Math; class Solution { public int minPathSum(int[][] grid) { int[][] dpStart = new int[grid.length + 1][grid[0].length + 1]; return dummySoln(dpStart,grid,grid[0].length-1, grid.length-1); } public int dummySoln(int[][] dp,int[][] grid, int x, int y) { if(dp[y][x] != 0){return dp[y][x];} if(x == y && x == 0) { return grid[0][0]; } int totalSum = grid[y][x]; int totalSumLeft = -1; int totalSumUp = -1; if(x-1 >= 0) { totalSumLeft = totalSum + dummySoln(dp, grid, x - 1, y); } if(y-1 >= 0) { totalSumUp = totalSum + dummySoln(dp, grid, x, y-1); } if(totalSumLeft == -1 && totalSumUp != -1) { return totalSumUp; } if(totalSumLeft != -1 && totalSumUp == -1) { return totalSumLeft; } dp[y][x] = Math.min(totalSumLeft, totalSumUp); return Math.min(totalSumLeft, totalSumUp); } } Problem solution in C++. class Solution { public: int minPathSum(vector<vector<int>>& grid) { if (grid.empty()) return 0; int n = grid.size(); int m = grid[0].size(); vector<vector<long long int>> res(n, vector<long long int>(m, 0)); res[0][0] = grid[0][0]; for(size_t i = 1; i < n; i++){ res[i][0] = res[i-1][0]+ grid[i][0]; } for(size_t i = 1; i < m; i++){ res[0][i] = res[0][i-1]+ grid[0][i]; } for(size_t i = 1; i < n; i++){ for(size_t j = 1; j < m; j++){ res[i][j] = min(res[i-1][j] + grid[i][j], res[i][j-1] + grid[i][j]); } } return res[n-1][m-1]; } }; Problem solution in C. int min(int a,int b){ if(a<b) return a; return b; } int minPathSum(int** grid, int gridSize, int* gridColSize){ int m=gridSize,n=gridColSize[0]; if(m==1 && n==1) return grid[0][0]; if(m==1){ int minn=0; for(int j=0;j<n;j++) minn+=grid[0][j]; return minn; } else{ int arr[m][n]; arr[0][0]=grid[0][0]; for(int j=0;j<n;j++){ for(int i=0;i<m;i++){ if(i==0 && j==0) continue; else if(i==0){ arr[i][j]=grid[i][j]+arr[i][j-1]; } else if(j==0) arr[i][0]=grid[i][0]+arr[i-1][j]; else if(i>0 && i<m){ arr[i][j]=grid[i][j]+min(arr[i][j-1],arr[i-1][j]); } } } return arr[m-1][n-1]; } } coding problems solutions