# 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;
}