Welcome to Subscribe On Youtube

2428. Maximum Sum of an Hourglass

Description

You are given an m x n integer matrix grid.

We define an hourglass as a part of the matrix with the following form:

Return the maximum sum of the elements of an hourglass.

Note that an hourglass cannot be rotated and must be entirely contained within the matrix.

 

Example 1:

Input: grid = [[6,2,1,3],[4,2,1,5],[9,2,8,7],[4,1,2,9]]
Output: 30
Explanation: The cells shown above represent the hourglass with the maximum sum: 6 + 2 + 1 + 2 + 9 + 2 + 8 = 30.

Example 2:

Input: grid = [[1,2,3],[4,5,6],[7,8,9]]
Output: 35
Explanation: There is only one hourglass in the matrix, with the sum: 1 + 2 + 3 + 5 + 7 + 8 + 9 = 35.

 

Constraints:

  • m == grid.length
  • n == grid[i].length
  • 3 <= m, n <= 150
  • 0 <= grid[i][j] <= 106

Solutions

Solution 1: Enumeration

We observe from the problem statement that each hourglass is a $3 \times 3$ matrix with the first and last elements of the middle row removed. Therefore, we can start from the top left corner, enumerate the middle coordinate $(i, j)$ of each hourglass, then calculate the sum of the elements in the hourglass, and take the maximum value.

The time complexity is $O(m \times n)$, where $m$ and $n$ are the number of rows and columns of the matrix, respectively. The space complexity is $O(1)$.

  • class Solution {
    public int maxSum(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        int ans = 0;
        for (int i = 1; i < m - 1; ++i) {
            for (int j = 1; j < n - 1; ++j) {
                int s = -grid[i][j - 1] - grid[i][j + 1];
                for (int x = i - 1; x <= i + 1; ++x) {
                    for (int y = j - 1; y <= j + 1; ++y) {
                        s += grid[x][y];
                    }
                }
                ans = Math.max(ans, s);
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    int maxSum(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        int ans = 0;
        for (int i = 1; i < m - 1; ++i) {
            for (int j = 1; j < n - 1; ++j) {
                int s = -grid[i][j - 1] - grid[i][j + 1];
                for (int x = i - 1; x <= i + 1; ++x) {
                    for (int y = j - 1; y <= j + 1; ++y) {
                        s += grid[x][y];
                    }
                }
                ans = max(ans, s);
            }
        }
        return ans;
    }
};

  • class Solution:
    def maxSum(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        ans = 0
        for i in range(1, m - 1):
            for j in range(1, n - 1):
                s = -grid[i][j - 1] - grid[i][j + 1]
                s += sum(
                    grid[x][y] for x in range(i - 1, i + 2) for y in range(j - 1, j + 2)
                )
                ans = max(ans, s)
        return ans


  • func maxSum(grid [][]int) (ans int) {
	m, n := len(grid), len(grid[0])
	for i := 1; i < m-1; i++ {
		for j := 1; j < n-1; j++ {
			s := -grid[i][j-1] - grid[i][j+1]
			for x := i - 1; x <= i+1; x++ {
				for y := j - 1; y <= j+1; y++ {
					s += grid[x][y]
				}
			}
			ans = max(ans, s)
		}
	}
	return
}

  • function maxSum(grid: number[][]): number {
    const m = grid.length;
    const n = grid[0].length;
    let ans = 0;
    for (let i = 1; i < m - 1; ++i) {
        for (let j = 1; j < n - 1; ++j) {
            let s = -grid[i][j - 1] - grid[i][j + 1];
            for (let x = i - 1; x <= i + 1; ++x) {
                for (let y = j - 1; y <= j + 1; ++y) {
                    s += grid[x][y];
                }
            }
            ans = Math.max(ans, s);
        }
    }
    return ans;
}

All Problems

All Solutions