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.lengthn == grid[i].length1 <= m, n <= 2000 <= 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.
