# 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

class Solution(object):
    def minFallingPathSum(self, A):
        """
        :type A: List[List[int]]
        :rtype: int
        """
        m, n = len(A), len(A[0])
        dp = [row[:] for row in A]

        for j in xrange(n):
            dp[0][j] = A[0][j]

        for i in xrange(1, m):
            for j in xrange(0, n):
                if j == 0:
                    dp[i][j] += min(dp[i - 1][j], dp[i - 1][j + 1])
                elif j == n - 1:
                    dp[i][j] += min(dp[i - 1][j - 1], dp[i - 1][j])
                else:
                    dp[i][j] += min(dp[i - 1][j - 1], dp[i - 1][j], dp[i - 1][j + 1])

        return min(dp[-1])


if __name__ == '__main__':
    print Solution().minFallingPathSum([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
