Hackerrank Java Solution: 2D Array - DS

Question: 

We define an hourglass in  to be a subset of values with indices falling in this pattern in 's graphical representation:

a b c
  d
e f g

There are  hourglasses in , and an hourglass sum is the sum of an hourglass' values. Calculate the hourglass sum for every hourglass in , then print the maximum hourglass sum.

For example, given the 2D array:

-9 -9 -9  1 1 1 
 0 -9  0  4 3 2
-9 -9 -9  1 2 3
 0  0  8  6 6 0
 0  0  0 -2 0 0
 0  0  1  2 4 0

We calculate the following  hourglass values:

-63, -34, -9, 12, 
-10, 0, 28, 23, 
-27, -11, -2, 10, 
9, 17, 25, 18

Our highest hourglass value is  from the hourglass:

0 4 3
  1
8 6 6

Solution:

The Solution is simple you just have to calculate the sum of the window or the hour glass and you would have to keep track of the previous maximum sum. 




For better understanding, here is the code.

 static int hourglassSum(int[][] arr) {

        int maxSum=-100;
        int r=0,s=0,p=0,q=0,sum=0,m=0,i=0,j=0,row=0,col=0,k=0;
        for(row=0;row<arr[0].length-2;row++){
            for(col=0;col<arr.length-2;col++)
            {
                i=arr[row][col];
                j=arr[row][col+1];
                k=arr[row][col+2];
                m=arr[row+1][col+1];
                p=arr[row+2][col];
                q=arr[row+2][col+1];
                r=arr[row+2][col+2];

                sum=p+q+r+i+j+k+m;
                if(sum>maxSum){
                    maxSum=sum;
                }
            }
        }
        System.out.print(maxSum);
        return maxSum;

    }

Thanks for reading.
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