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.lengthn == grid[i].length1 <= m, n <= 2000 <= 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 × n), and the space complexity is O(m × n). Here, m and n are the number of rows and columns of the grid, respectively.
PythonJavaC++GoTypeScriptRustJavaScriptC#
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](code-box)
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];
}
}(code-box)
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];
}
};(code-box)
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]
}(code-box)
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];
}(code-box)
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]
}
}(code-box)
/**
* @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];
};(code-box)
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];
}
}(code-box)
