Minimum Falling Path Sum

Given a square array of integers A, we want the minimum sum of a falling path through A.

A falling path starts at any element in the first row, and chooses one element from each row. The next row's choice must be in a column that is different from the previous row's column by at most one.

Example 1:

Input: [[1,2,3],[4,5,6],[7,8,9]]
Output: 12
Explanation: 
The possible falling paths are:

• [1,4,7], [1,4,8], [1,5,7], [1,5,8], [1,5,9]

• [2,4,7], [2,4,8], [2,5,7], [2,5,8], [2,5,9], [2,6,8], [2,6,9]

• [3,5,7], [3,5,8], [3,5,9], [3,6,8], [3,6,9]

The falling path with the smallest sum is [1,4,7], so the answer is 12.

Note:

1. 1 <= A.length == A[0].length <= 100

2. -100 <= A[i][j] <= 100

class Solution {
    public int minFallingPathSum(int[][] A) {
        // dp[i][j] shows min fall path from this element
        int[][] dp = new int[A.length][A.length];
        // bottum up approach
        for (int i = A.length - 1; i >= 0; i--) {
            for (int j = 0; j < A.length; j++) {
                dp[i][j] = A[i][j];
                if(i==A.length-1)
                    continue; // base case
                if (j == 0)
                    dp[i][j] += Math.min(dp[i + 1][j], dp[i + 1][j + 1]);
                else if (j == A.length - 1)
                    dp[i][j] += Math.min(dp[i + 1][j - 1], dp[i + 1][j]);
                else
                    dp[i][j] += Math.min(dp[i + 1][j], Math.min(dp[i + 1][j - 1], dp[i + 1][j + 1]));
            }
        }
        // finding min from first row
        int min = Integer.MAX_VALUE;
        for (int x : dp[0])
            min = Math.min(x, min);
        return min;
    }
}