C++11.

We need to:

Traverse a 6x6 2D array.

For each possible “hourglass” (shape of 7 cells), calculate its sum.

Track and return the maximum sum.

Here’s a clean solution:

include

using namespace std;

int hourglassSum(vector> arr) { int maxSum = INT_MIN; // since values can be negative

for (int i = 0; i <= 3; i++) {
    for (int j = 0; j <= 3; j++) {
        int sum = arr[i][j] + arr[i][j+1] + arr[i][j+2]   // top row
                + arr[i+1][j+1]                           // middle
                + arr[i+2][j] + arr[i+2][j+1] + arr[i+2][j+2]; // bottom row

        maxSum = max(maxSum, sum);
    }
}
return maxSum;

}

int main() { vector> arr(6, vector(6)); for (int i = 0; i < 6; i++) { for (int j = 0; j < 6; j++) { cin >> arr[i][j]; } } cout << hourglassSum(arr) << endl; return 0; }

✅ Explanation:

We only go up to index 3 (not 5), since an hourglass needs 3 rows and 3 columns.

At each (i, j), we compute the hourglass sum.

Keep updating maxSum.

Return the largest sum found.

🔹 For the sample input in your problem, this outputs 19.