My python sollution`

def hourglassSum(a):
    ma = -float('inf')
    for i in range(1,5):
        for j in range(1,5):
            top = bottom = 0
            for dy in [-1,0,1] :
                top += a[i-1][j+dy]
                bottom += a[i+1][j+dy]
            ma = max(ma,top + a[i][j] + bottom)
    return ma

A solution in Python that respects the constraints proposed by the challenge (and focuses on readability):

"""
Constraints:
    -9 <= arr[i][j] <= 9
    0 <= i,j <= 5
"""

def get_hour_glass_sum(arr: List[int], x: int, y: int) -> int:
    first_line = sum([arr[x][y + i] for i in range(3)])
    second_line = arr[x + 1][y + 1]
    third_line = sum([arr[x + 2][y + i] for i in range(3)])
    
    return first_line + second_line + third_line

def hourglassSum(arr):
    greater_sum = None
    
    for x in range(4):
        for y in range(4):
            current_sum = get_hour_glass_sum(arr, x, y)
            
            if greater_sum is None:
                greater_sum = current_sum
                continue
            
            greater_sum = max(current_sum, greater_sum)
            
    return greater_sum

this is my solution :

**My Code In Java: **

int temporary = Integer.MIN_VALUE;
        for( int i = 0; i < arr.size() - 2; i++){
            for( int j = 0 ; j < arr.size() - 2; j++){
                int a = arr.get(i).get(j);
                int b = arr.get(i).get(j+1);
                int c = arr.get(i).get(j+2);
                int d = arr.get(i+1).get(j+1);
                int e = arr.get(i+2).get(j);
                int f = arr.get(i+2).get(j+1);
                int g = arr.get(i+2).get(j+2);
                
                int result = a + b + c + d + e + f + g;
                if (result > temporary){
                    temporary = result;
                }
            }
        }
        
        return temporary;

Here is the solution in C language. (Paste it in VS Code or format it (can also use gpt to format).

include

void input(int arr[6][6]) { for(int row = 0; row < 6; row++) { for(int col = 0; col < 6; col++) { scanf("%d", &arr[row][col]); } } } int highest_sum(int arr[6][6]) { // Basically, hourglass format is like. // These all alphabets are basically represent columns.
// a1, a2, a3 // b1 // c1, c2, c3 int result_sum = -100, temp_sum = 0; // we have initialized result_sum with -100 not 0 because if the numbers are negative and first sum we get is also negative, so it will not get stored becuase of if condition. for(int row=0; row<4; row++) { for(int cols=0; cols<4; cols++) {
temp_sum = arr[row][cols] + arr[row][cols+1] + arr[row][cols+2] + arr[row+1][cols+1] + arr[row+2][cols] + arr[row+2][cols+1] + arr[row+2][cols+2] ; //a1 + a2 + a3 + b1 +c1 + c2 + c3 if (temp_sum>result_sum) { // so that the bigger sum gets stored in the result. result_sum = temp_sum; } } } return result_sum; } int main() { int matrix[6][6]; int highest_hourglass_sum; input(matrix); //input function is created. highest_hourglass_sum = highest_sum(matrix); printf("%d",highest_hourglass_sum); return 0; } int main() { int matrix[6][6]; int highest_hourglass_sum; input(matrix); //input function is created. highest_hourglass_sum = highest_sum(matrix); printf("%d",highest_hourglass_sum); return 0; }

Your method of breaking down the 6x6 matrix into manageable 6x3 sections and using loops to calculate hourglass sums is a smart, structured solution—especially for fixed sizes. It's similar to how in Bowmasters, you might focus on specific strategies to collect gems efficiently within a known game mode or map. However, just like your matrix approach might not scale well for NxM matrices, a rigid strategy in Bowmasters might not adapt well across different challenges or updates. Flexibility is key in both cases for long-term success and efficiency.