Question:

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 <= A.length == A[0].length <= 100
  • -100 <= A[i][j] <= 100

Solution:

  • Method 1: DP O(MN) AC 96.21% 
      class Solution {
      public int minFallingPathSum(int[][] A) {
          int len = A.length;
          if(len == 0) return 0;
          int[][] dp = new int[len][len];
          for(int i = 0; i < len; i++){
              dp[0][i] = A[0][i];
          }
          for(int i = 1; i < len; i++){
              for(int j = 0; j < len; j++){
                  dp[i][j] = Math.min(Math.min(j > 0 ? dp[i - 1][j - 1]: Integer.MAX_VALUE, j + 1 < len ? dp[i - 1][j + 1]: Integer.MAX_VALUE), dp[i - 1][j]) + A[i][j];
              }
          }
          int res = Integer.MAX_VALUE;
          for(int i = 0; i < len; i++)
              res = Math.min(res, dp[len - 1][i]);
          return res;
      }
  }