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).