Minimum falling path sum

class Solution:
    def minFallingPathSum(self, matrix: List[List[int]]) -> int:
        m, n = len(matrix), len(matrix[0])
        dp = [[0] * n for _ in range(m)]
        for i in range(n):
            dp[0][i] = matrix[0][i]

        for i in range(1, m):
            for j in range(n):
                if j == 0:
                    dp[i][j] = min(dp[i - 1][j], dp[i - 1][j + 1]) + matrix[i][j]
                elif j == n - 1:
                    dp[i][j] = min(dp[i - 1][j], dp[i - 1][j - 1]) + matrix[i][j]
                else:
                    dp[i][j] = min(dp[i - 1][j], dp[i - 1][j - 1], dp[i - 1][j + 1]) + matrix[i][j]
        # print(dp)
        return min(dp[-1])

Minimum Falling Path Sum

Difficulty: Medium

Given an n x n array of integers matrix, return the minimum sum of any falling path through matrix.

A falling path starts at any element in the first row and chooses the element in the next row that is either directly below or diagonally left/right. Specifically, the next element from position (row, col) will be (row + 1, col - 1), (row + 1, col), or (row + 1, col + 1).

 

Example 1:

Input: matrix = [[2,1,3],[6,5,4],[7,8,9]]
Output: 13
Explanation: There are two falling paths with a minimum sum as shown.

Example 2:

Input: matrix = [[-19,57],[-40,-5]]
Output: -59
Explanation: The falling path with a minimum sum is shown.

 

Constraints:

  • n == matrix.length == matrix[i].length
  • 1 <= n <= 100
  • -100 <= matrix[i][j] <= 100