# 64. Minimum Path Sum

## Description

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] <= 200

## Solutions

Solution 1: Dynamic Programming

We define $f[i][j]$ to represent the minimum path sum from the top left corner to $(i, j)$. Initially, $f[0][0] = grid[0][0]$, and the answer is $f[m - 1][n - 1]$.

Consider $f[i][j]$:

• If $j = 0$, then $f[i][j] = f[i - 1][j] + grid[i][j]$;
• If $i = 0$, then $f[i][j] = f[i][j - 1] + grid[i][j]$;
• If $i > 0$ and $j > 0$, then $f[i][j] = \min(f[i - 1][j], f[i][j - 1]) + grid[i][j]$.

Finally, return $f[m - 1][n - 1]$.

The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the number of rows and columns of the grid, respectively.

• class Solution {
public int minPathSum(int[][] grid) {
int m = grid.length, n = grid[0].length;
int[][] f = new int[m][n];
f[0][0] = grid[0][0];
for (int i = 1; i < m; ++i) {
f[i][0] = f[i - 1][0] + grid[i][0];
}
for (int j = 1; j < n; ++j) {
f[0][j] = f[0][j - 1] + grid[0][j];
}
for (int i = 1; i < m; ++i) {
for (int j = 1; j < n; ++j) {
f[i][j] = Math.min(f[i - 1][j], f[i][j - 1]) + grid[i][j];
}
}
return f[m - 1][n - 1];
}
}

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

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


• func minPathSum(grid [][]int) int {
m, n := len(grid), len(grid[0])
f := make([][]int, m)
for i := range f {
f[i] = make([]int, n)
}
f[0][0] = grid[0][0]
for i := 1; i < m; i++ {
f[i][0] = f[i-1][0] + grid[i][0]
}
for j := 1; j < n; j++ {
f[0][j] = f[0][j-1] + grid[0][j]
}
for i := 1; i < m; i++ {
for j := 1; j < n; j++ {
f[i][j] = min(f[i-1][j], f[i][j-1]) + grid[i][j]
}
}
return f[m-1][n-1]
}

• function minPathSum(grid: number[][]): number {
const m = grid.length;
const n = grid[0].length;
const f: number[][] = Array(m)
.fill(0)
.map(() => Array(n).fill(0));
f[0][0] = grid[0][0];
for (let i = 1; i < m; ++i) {
f[i][0] = f[i - 1][0] + grid[i][0];
}
for (let j = 1; j < n; ++j) {
f[0][j] = f[0][j - 1] + grid[0][j];
}
for (let i = 1; i < m; ++i) {
for (let j = 1; j < n; ++j) {
f[i][j] = Math.min(f[i - 1][j], f[i][j - 1]) + grid[i][j];
}
}
return f[m - 1][n - 1];
}


• /**
* @param {number[][]} grid
* @return {number}
*/
var minPathSum = function (grid) {
const m = grid.length;
const n = grid[0].length;
const f = Array(m)
.fill(0)
.map(() => Array(n).fill(0));
f[0][0] = grid[0][0];
for (let i = 1; i < m; ++i) {
f[i][0] = f[i - 1][0] + grid[i][0];
}
for (let j = 1; j < n; ++j) {
f[0][j] = f[0][j - 1] + grid[0][j];
}
for (let i = 1; i < m; ++i) {
for (let j = 1; j < n; ++j) {
f[i][j] = Math.min(f[i - 1][j], f[i][j - 1]) + grid[i][j];
}
}
return f[m - 1][n - 1];
};


• public class Solution {
public int MinPathSum(int[][] grid) {
int m = grid.Length, n = grid[0].Length;
int[,] f = new int[m, n];
f[0, 0] = grid[0][0];
for (int i = 1; i < m; ++i) {
f[i, 0] = f[i - 1, 0] + grid[i][0];
}
for (int j = 1; j < n; ++j) {
f[0, j] = f[0, j - 1] + grid[0][j];
}
for (int i = 1; i < m; ++i) {
for (int j = 1; j < n; ++j) {
f[i, j] = Math.Min(f[i - 1, j], f[i, j - 1]) + grid[i][j];
}
}
return f[m - 1, n - 1];
}
}

• impl Solution {
pub fn min_path_sum(mut grid: Vec<Vec<i32>>) -> i32 {
let m = grid.len();
let n = grid[0].len();
for i in 1..m {
grid[i][0] += grid[i - 1][0];
}
for i in 1..n {
grid[0][i] += grid[0][i - 1];
}
for i in 1..m {
for j in 1..n {
grid[i][j] += grid[i][j - 1].min(grid[i - 1][j]);
}
}
grid[m - 1][n - 1]
}
}