leetcode - 931. Minimum Falling Path Sum

Description

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

Solution

Simple dp. The transformation equation is:
$$dp[i][j]=\min (dp[i+1][j-1],dp[i+1][j],dp[i+1][j+1])$$

Time complexity: $o(n^2)$
Space complexity: $o(n^2)$, could be reduced to $o(n)$

Code

class Solution:
    def minFallingPathSum(self, matrix: List[List[int]]) -> int:
        n = len(matrix)
        dp = [[100] * n for _ in range(n)]
        dp[-1] = matrix[-1].copy()
        max_val = 1000001
        for i in range(n - 2, -1, -1):
            for j in range(n):
                if j - 1 >= 0:
                    left_val = dp[i + 1][j - 1]
                else:
                    left_val = max_val
                if j + 1 < n:
                    right_val = dp[i + 1][j + 1]
                else:
                    right_val = max_val
                dp[i][j] = matrix[i][j] + min([dp[i + 1][j], left_val, right_val])
        return min(dp[0])