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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.

  • 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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