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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

• Java
• C++
• Python
• Go

• 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;
      }
  }
  

• 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);
      }
  };
  

• 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])
  
  

• 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 {
  					x = min(x, f[i-1][k])
  				}
  			}
  			if x == inf {
  				x = 0
  			}
  			f[i][j] = v + x
  		}
  	}
  	return slices.Min(f[n])
  }
  

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