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1289. Minimum Falling Path Sum II

Description

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

Solution 1

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class Solution:
    def minFallingPathSum(self, grid: List[List[int]]) -> int:
        n = len(grid)
        f = [[0] * n for _ in range(n + 1)]
        for i, row in enumerate(grid, 1):
            for j, v in enumerate(row):
                x = min((f[i - 1][k] for k in range(n) if k != j), default=0)
                f[i][j] = v + x
        return min(f[n])

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class Solution {
    public int minFallingPathSum(int[][] grid) {
        int n = grid.length;
        int[][] f = new int[n + 1][n];
        final int inf = 1 << 30;
        for (int i = 1; i <= n; ++i) {
            for (int j = 0; j < n; ++j) {
                int x = inf;
                for (int k = 0; k < n; ++k) {
                    if (k != j) {
                        x = Math.min(x, f[i - 1][k]);
                    }
                }
                f[i][j] = grid[i - 1][j] + (x == inf ? 0 : x);
            }
        }
        int ans = inf;
        for (int x : f[n]) {
            ans = Math.min(ans, x);
        }
        return ans;
    }
}

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class Solution {
public:
    int minFallingPathSum(vector<vector<int>>& grid) {
        int n = grid.size();
        int f[n + 1][n];
        memset(f, 0, sizeof(f));
        const int inf = 1 << 30;
        for (int i = 1; i <= n; ++i) {
            for (int j = 0; j < n; ++j) {
                int x = inf;
                for (int k = 0; k < n; ++k) {
                    if (k != j) {
                        x = min(x, f[i - 1][k]);
                    }
                }
                f[i][j] = grid[i - 1][j] + (x == inf ? 0 : x);
            }
        }
        return *min_element(f[n], f[n] + n);
    }
};

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func minFallingPathSum(grid [][]int) int {
    n := len(grid)
    f := make([][]int, n+1)
    for i := range f {
        f[i] = make([]int, n)
    }
    const inf = 1 << 30
    for i, row := range grid {
        i++
        for j, v := range row {
            x := inf
            for k := range row {
                if k != j