README_EN.md

931. Minimum Falling Path Sum

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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

Solutions

Dynamic programming.

Python3

class Solution:
    def minFallingPathSum(self, matrix: List[List[int]]) -> int:
        n = len(matrix)
        for i in range(1, n):
            for j in range(n):
                mi = matrix[i - 1][j]
                if j > 0:
                    mi = min(mi, matrix[i - 1][j - 1])
                if j < n - 1:
                    mi = min(mi, matrix[i - 1][j + 1])
                matrix[i][j] += mi
        return min(matrix[n - 1])

Java

class Solution {
    public int minFallingPathSum(int[][] matrix) {
        int n = matrix.length;
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                int mi = matrix[i - 1][j];
                if (j > 0) {
                    mi = Math.min(mi, matrix[i - 1][j - 1]);
                }
                if (j < n - 1) {
                    mi = Math.min(mi, matrix[i - 1][j + 1]);
                }
                matrix[i][j] += mi;
            }
        }
        int res = Integer.MAX_VALUE;
        for (int j = 0; j < n; ++j) {
            res = Math.min(res, matrix[n - 1][j]);
        }
        return res;
    }
}

C++

class Solution {
public:
    int minFallingPathSum(vector<vector<int>>& matrix) {
        int n = matrix.size();
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                int mi = matrix[i - 1][j];
                if (j > 0) mi = min(mi, matrix[i - 1][j - 1]);
                if (j < n - 1) mi = min(mi, matrix[i - 1][j + 1]);
                matrix[i][j] += mi;
            }
        }
        int res = INT_MAX;
        for (int j = 0; j < n; ++j) {
            res = min(res, matrix[n - 1][j]);
        }
        return res;
    }
};

Go

func minFallingPathSum(matrix [][]int) int {
    n := len(matrix)
    for i := 1; i < n; i++ {
        for j := 0; j < n; j++ {
            mi := matrix[i - 1][j]
            if j > 0 && mi > matrix[i - 1][j - 1] {
                mi = matrix[i - 1][j - 1]
            }
            if j < n - 1 && mi > matrix[i - 1][j + 1] {
                mi = matrix[i - 1][j + 1]
            }
            matrix[i][j] += mi
        }
    }
    res := 10000
    for j := 0; j < n; j++ {
        if res > matrix[n - 1][j] {
            res = matrix[n - 1][j]
        }
    }
    return res
}

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