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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

Solutions

Dynamic programming.

• Java
• C++
• Python
• Go
• TypeScript

• class Solution {
      public int minFallingPathSum(int[][] matrix) {
          int n = matrix.length;
          var f = new int[n];
          for (var row : matrix) {
              var g = f.clone();
              for (int j = 0; j < n; ++j) {
                  if (j > 0) {
                      g[j] = Math.min(g[j], f[j - 1]);
                  }
                  if (j + 1 < n) {
                      g[j] = Math.min(g[j], f[j + 1]);
                  }
                  g[j] += row[j];
              }
              f = g;
          }
          // return Arrays.stream(f).min().getAsInt();
          int ans = 1 << 30;
          for (int x : f) {
              ans = Math.min(ans, x);
          }
          return ans;
      }
  }
  

• class Solution {
  public:
      int minFallingPathSum(vector<vector<int>>& matrix) {
          int n = matrix.size();
          vector<int> f(n);
          for (auto& row : matrix) {
              auto g = f;
              for (int j = 0; j < n; ++j) {
                  if (j) {
                      g[j] = min(g[j], f[j - 1]);
                  }
                  if (j + 1 < n) {
                      g[j] = min(g[j], f[j + 1]);
                  }
                  g[j] += row[j];
              }
              f = move(g);
          }
          return *min_element(f.begin(), f.end());
      }
  };
  

• class Solution:
      def minFallingPathSum(self, matrix: List[List[int]]) -> int:
          n = len(matrix)
          f = [0] * n
          for row in matrix:
              g = [0] * n
              for j, x in enumerate(row):
                  l, r = max(0, j - 1), min(n, j + 2)
                  g[j] = min(f[l:r]) + x
              f = g
          return min(f)
  
  

• func minFallingPathSum(matrix [][]int) int {
  	n := len(matrix)
  	f := make([]int, n)
  	for _, row := range matrix {
  		g := make([]int, n)
  		copy(g, f)
  		for j, x := range row {
  			if j > 0 {
  				g[j] = min(g[j], f[j-1])
  			}
  			if j+1 < n {
  				g[j] = min(g[j], f[j+1])
  			}
  			g[j] += x
  		}
  		f = g
  	}
  	return slices.Min(f)
  }
  

• function minFallingPathSum(matrix: number[][]): number {
      const n = matrix.length;
      const f: number[] = new Array(n).fill(0);
      for (const row of matrix) {
          const g = f.slice();
          for (let j = 0; j < n; ++j) {
              if (j > 0) {
                  g[j] = Math.min(g[j], f[j - 1]);
              }
              if (j + 1 < n) {
                  g[j] = Math.min(g[j], f[j + 1]);
              }
              g[j] += row[j];
          }
          f.splice(0, n, ...g);
      }
      return Math.min(...f);
  }
  
  

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