# Minimum Path Sum

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:**

```
Input:
[
  [1,3,1],
  [1,5,1],
  [4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.
```

## Analysis

与 Unique Paths很相似，不过路径有了权重，因此在初始化和状态转移方程上稍有区别：

**状态**：`dp[i][j]` - 从起点到达(i, j)的最小路径和Min Path Sum\
**状态转移方程**: `dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];` - 左侧`(i, j - 1)`和上方 `(i - 1, j)`位置的路径和较小值，加上`(i, j)` 位置的权重\
**初始条件**: `dp[0][0] = grid[0][0]`, `dp[i][0] = dp[i - 1][0] + grid[i][0]; (i = 0, ... m - 1)`, `dp[0][j] = dp[0][j - 1] + grid[0][j]; (j = 0, ..., n - 1)`

**答案**：`dp[m - 1][n - 1]` 即终点位置

## Solution

DP - O(mn) space, O(mn) time (4 \~ 6ms 51.84% AC)

```java
class Solution {
    public int minPathSum(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0) return 0;
        int m = grid.length;
        int n = grid[0].length;
        int[][] dp = new int[m][n];
        dp[0][0] = grid[0][0];
        for (int i = 1; i < m; i++) {
            dp[i][0] = dp[i - 1][0] + grid[i][0];
        }
        for (int j = 1; j < n; j++) {
            dp[0][j] = dp[0][j - 1] + grid[0][j];
        }
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[i][j] = Math.min(dp[i][j - 1], dp[i - 1][j]) + grid[i][j];
            }
        }
        return dp[m - 1][n - 1];

    }
}
```

DP - without extra space (reusing grid\[]\[] array itself)

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