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.
 
Constraints:

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

Difficulty: interview

Starter code:

class Solution:
    def minFallingPathSum(self, A: List[List[int]]) -> int:
        

Solutions:
class Solution:
    def minFallingPathSum(self, A: List[List[int]]) -> int:
        dp = [A[0][:], [0 for _ in A[0]]]
        for i in range(1, len(A)):
            for j in range(len(A[i])):
                dp[i & 1][j] = min([dp[(i - 1) & 1][j + k] for k in (-1, 0, 1) if 0 <= j + k < len(A[i])]) + A[i][j]
        return min(dp[(len(A) - 1) & 1])


class Solution:
    def minFallingPathSum(self, A: List[List[int]]) -> int:
        dp = [A[0][:], [0 for _ in A[0]]]
        for i in range(1, len(A)):
            for j in range(len(A[i])):
                dp[i & 1][j] = min([dp[(i - 1) & 1][j + k] + A[i][j] for k in (-1, 0, 1) if 0 <= j + k < len(A[i])])
        return min(dp[(len(A) - 1) & 1])