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2428. Maximum Sum of an Hourglass 👍

  • Time: $O(mn)$
  • Space: $O(1)$
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class Solution {
 public:
  int maxSum(vector<vector<int>>& grid) {
    int ans = 0;

    for (int i = 1; i + 1 < grid.size(); ++i)
      for (int j = 1; j + 1 < grid[0].size(); ++j)
        ans =
            max(ans, grid[i - 1][j - 1] + grid[i - 1][j] + grid[i - 1][j + 1] +
                         grid[i][j] + grid[i + 1][j - 1] + grid[i + 1][j] +
                         grid[i + 1][j + 1]);

    return ans;
  }
};

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public class Solution {
  public int maxSum(int[][] grid) {
    int ans = 0;

    for (int i = 1; i + 1 < grid.length; ++i)
      for (int j = 1; j + 1 < grid[0].length; ++j)
        ans = Math.max(ans, grid[i - 1][j - 1] + grid[i - 1][j] + grid[i - 1][j + 1] + grid[i][j] +
                                grid[i + 1][j - 1] + grid[i + 1][j] + grid[i + 1][j + 1]);

    return ans;
  }
}

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class Solution:
  def maxSum(self, grid: List[List[int]]) -> int:
    return max(grid[i - 1][j - 1] + grid[i - 1][j] + grid[i - 1][j + 1] + grid[i][j] +
               grid[i + 1][j - 1] + grid[i + 1][j] + grid[i + 1][j + 1]
               for i in range(1, len(grid) - 1)
               for j in range(1, len(grid[0]) - 1))