2428 Maximum Sum of an Hourglass

We just brute force our way into the matrix, do bound checks for our values and just calculate the required hour glass sum, and finally update our max sum with every iteration.

Code:

Java

class Solution {
    public int maxSum(int[][] grid) {
        int max = Integer.MIN_VALUE;
        for (int i=0;i<grid.length;++i) {
            for (int j=0;j<grid[i].length;++j) {
                if (i == 0 || i >= grid.length-1 || j == 0 || j >= grid[i].length-1) continue;
 
                int sum = grid[i][j] + 
                        grid[i-1][j-1] +
                        grid[i-1][j+1] + 
                        grid[i-1][j] + 
                        grid[i+1][j] + 
                        grid[i+1][j-1] +
                        grid[i+1][j+1];
                max = Math.max(max,sum);
            }
        }
        return max;
    }
}

Big O Analysis

• Runtime

  The runtime complexity here is $O(N^2)$ .

• Memory

  The memory usage is O(1) .

— A

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