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Formatted question description: https://leetcode.ca/all/2428.html

2428. Maximum Sum of an Hourglass

  • Difficulty: Medium.
  • Related Topics: .
  • Similar Questions: Matrix Block Sum.

Problem

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] <= 10^6

Solution (Java, C++, Python)

  • 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 t = 0;
                for (int x = i - 1; x <= i + 1; ++x) {
                    for (int y = j - 1; y <= j + 1; ++y) {
                        t += grid[x][y];
                    }
                }
                t -= grid[i][j - 1];
                t -= grid[i][j + 1];
                ans = Math.max(ans, t);
            }
        }
        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 t = 0;
                for (int x = i - 1; x <= i + 1; ++x) {
                    for (int y = j - 1; y <= j + 1; ++y) {
                        t += grid[x][y];
                    }
                }
                t -= grid[i][j - 1];
                t -= grid[i][j + 1];
                ans = max(ans, t);
            }
        }
        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):
                t = 0
                for x in [i - 1, i, i + 1]:
                    for y in [j - 1, j, j + 1]:
                        t += grid[x][y]

                t -= grid[i][j - 1]
                t -= grid[i][j + 1]
                ans = max(ans, t)
        return ans


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

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

  • function maxSum(grid: number[][]): number {
    const m = grid.length,
        n = grid[0].length;
    let threeSum = Array.from({ length: m }, () => new Array(n - 2).fill(0));
    for (let i = 0; i < m; i++) {
        for (let j = 1; j < n - 1; j++) {
            threeSum[i][j - 1] = grid[i][j - 1] + grid[i][j] + grid[i][j + 1];
        }
    }
    let ans = 0;
    for (let i = 1; i < m - 1; i++) {
        for (let j = 1; j < n - 1; j++) {
            ans = Math.max(
                ans,
                threeSum[i - 1][j - 1] + grid[i][j] + threeSum[i + 1][j - 1],
            );
        }
    }
    return ans;
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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