README.md

1289. Minimum Falling Path Sum II

Given an n x n integer matrix grid, return the minimum sum of a falling path with non-zero shifts.

A falling path with non-zero shifts is a choice of exactly one element from each row of grid such that no two elements chosen in adjacent rows are in the same column.

Example 1:

Input: grid = [[1,2,3],[4,5,6],[7,8,9]]
Output: 13
Explanation:
The possible falling paths are:
[1,5,9], [1,5,7], [1,6,7], [1,6,8],
[2,4,8], [2,4,9], [2,6,7], [2,6,8],
[3,4,8], [3,4,9], [3,5,7], [3,5,9]
The falling path with the smallest sum is [1,5,7], so the answer is 13.

Example 2:

Input: grid = [[7]]
Output: 7

Constraints:

• n == grid.length == grid[i].length
• 1 <= n <= 200
• -99 <= grid[i][j] <= 99

Solutions (Rust)

1. Solution

impl Solution {
    pub fn min_falling_path_sum(mut grid: Vec<Vec<i32>>) -> i32 {
        if grid.len() == 1 {
            return grid[0][0];
        }

        let mut min_col2 = [grid.len(); 2];
        let mut min_sum2 = [0; 2];

        for r in 0..grid.len() {
            let mut i = grid.len();
            let mut j = grid.len();

            for c in 0..grid.len() {
                if c != min_col2[0] {
                    grid[r][c] += min_sum2[0];
                } else {
                    grid[r][c] += min_sum2[1];
                }

                if grid[r][c] <= *grid[r].get(i).unwrap_or(&i32::MAX) {
                    j = i;
                    i = c;
                } else if grid[r][c] < *grid[r].get(j).unwrap_or(&i32::MAX) {
                    j = c;
                }
            }

            min_col2 = [i, j];
            min_sum2 = [grid[r][i], grid[r][j]];
        }

        min_sum2[0].min(min_sum2[1])
    }
}