Different and Simple approach. code in C++. Logic: - First we find all the possible 3X3 matrices and their sum - Then we subtract those particular two elements from the 3X3 matrix sum so that we get sum of hourglass indirectly. - And compare them for maximum sum.

Et voila! Sometimes we have to look at the problem with a different perspective.

My Code:

      int main(){
      vector< vector<int> > arr(6,vector<int>(6));
      for(int arr_i = 0;arr_i < 6;arr_i++){
      for(int arr_j = 0;arr_j < 6;arr_j++){
      cin >> arr[arr_i][arr_j];
      }
      }
      int sumhour=-1000,sum=0; //sumhour intialized for      negative value so to calculate sum for negative numbers
      for(int iog=0;iog<4;iog++) //for loop to generate first element of all 3X3 matrices possible
     {
     for(int jog=0;jog<4;jog++) //for loop to generate first element of all 3X3 matrices possible
     {
        sum=0;
        for(int i=iog;i<iog+3;i++) //to access all other members of 3X3 matrix
        {
            for(int j=jog;j<jog+3;j++) //to access all other members of 3X3 matrix
            { 

                sum=sum+arr[i][j];          //sum of 3X3 matrix                     
             }
          }
           sum=sum-arr[iog+1][jog]-arr[iog+1][jog+2]; //here we subtract those two elements from 3X3 which are not included in hourglass

         if(sum>sumhour) sumhour=sum;

         }
      }

     cout<<sumhour;




    return 0;
 }