Min Sum Path in Matrix

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.

Input : 

    [  1 3 2
       4 3 1
       5 6 1
    ]

Output : 8
     1 -> 3 -> 2 -> 1 -> 1

Solution:

Time/Space: O(n^2)

public class Solution {
    public int minPathSum(int[][] A) {
        // dp[i][j] = min path sum so far
        // dp[i][j] = A[i][j] + min(dp[i - 1][j], dp[i][j - 1])
        // dp[0][0] = A[0][0]
        // dp[m - 1][n - 1]
        int m = A.length;
        int n = A[0].length;
        int min = A[0][0];
        int[][] dp = new int[m][n];
        dp[0][0] = A[0][0];
        for (int j = 1; j < n; j ++) {
            dp[0][j] = dp[0][j - 1] + A[0][j];
        }
        for (int i = 1; i < m; i ++) {
            dp[i][0] = dp[i - 1][0] + A[i][0];
        }
        for (int i = 1; i < m; i ++) {
            for (int j = 1; j < n; j ++) {
                dp[i][j] = A[i][j] + Math.min(dp[i - 1][j], dp[i][j - 1]);
            }
        }
        return dp[m - 1][n - 1];
    }
}

Space optimized:

public class Solution {
    public int minPathSum(int[][] A) {
        // dp[i][j] = min path sum so far
        // dp[i][j] = A[i][j] + min(dp[i - 1][j], dp[i][j - 1])
        // dp[0][0] = A[0][0]
        // dp[m - 1][n - 1]
        int m = A.length;
        int n = A[0].length;
        int[] dp = new int[n];
        dp[0] = A[0][0];
        for (int j = 1; j < n; j ++) {
            dp[j] = dp[j - 1] + A[0][j];
        }
        for (int i = 1; i < m; i ++) {
            dp[0] = dp[0] + A[i][0];
            for (int j = 1; j < n; j ++) {
                dp[j] = A[i][j] + Math.min(dp[j], dp[j - 1]);
            }
        }
        return dp[n - 1];
    }
}