Easy to read JAVA solution:

public static int hourglassSum(List<List<Integer>> arr) {

    int arrLength = 6;
    int subArrLength = 6;
    int hourGlassHeight = 3;
    int maxSum = Integer.MIN_VALUE;
    int lastTopRow = arrLength - hourGlassHeight;

    for (int i = 0; i <= lastTopRow; i++) {
        for (int topCenter = 1; topCenter < subArrLength - 1; topCenter++) {
            List<Integer> topRow = arr.get(i);
            List<Integer> midRow = arr.get(i + 1);
            List<Integer> botRow = arr.get(i + 2);

            int topRowSum = (
                topRow.get(topCenter - 1) +
                topRow.get(topCenter) +
                topRow.get(topCenter + 1)
            );
            int midRowSum = midRow.get(topCenter);
            int botRowSum = (
                botRow.get(topCenter - 1) +
                botRow.get(topCenter) +
                botRow.get(topCenter + 1)
            );
            int currentSum = topRowSum + midRowSum + botRowSum;

            if (currentSum > maxSum) maxSum = currentSum;
        }
    }
    return maxSum;
}

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.

max_sum = -float("inf")

for i in range(4):
    for j in range(4):
        top = arr[i][j]+arr[i][j+1]+arr[i][j+2]
        middle = arr[i+1][j+1]
        bottom = arr[i+2][j]+arr[i+2][j+1]+arr[i+2][j+2]
        total = top + middle + bottom
        max_sum = max(max_sum, total)
return max_sum

The 2D Array – DS problem tests how you handle multi-dimensional arrays efficiently, especially for pattern extraction like hourglass sums. In the same way, planning an Umrah from Montréal requires organizing multiple elements (flights, hotels, guidance) in a structured way, ensuring everything fits together smoothly for the best experience.The 2D Array – DS problem tests how you handle multi-dimensional arrays efficiently, especially for pattern extraction like hourglass sums. In the same way, planning an Umrah from Montréal requires organizing multiple elements (flights, hotels, guidance) in a structured way, ensuring everything fits together smoothly for the best experience.