HourGlass.java

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;

public class Solution {



    private static final Scanner scanner = new Scanner(System.in);

    public static void main(String[] args) {
        int[][] arr = new int[6][6];

        for (int i = 0; i < 6; i++) {
            String[] arrRowItems = scanner.nextLine().split(" ");
            scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

            for (int j = 0; j < 6; j++) {
                int arrItem = Integer.parseInt(arrRowItems[j]);
                arr[i][j] = arrItem;
            }
        }
        int large=-9,sum;
        for(int i=0;i<4;i++)
        {
            for(int j=0;j<4;j++)
            {
                
                
                    sum = arr[i][j]+arr[i][j+1]+arr[i][j+2]+arr[i+1][j+1]+arr[i+2][j]+arr[i+2][j+1]+arr[i+2][j+2];
                if(sum>large)
                {
                    large=sum;
                }
                
            }
        }
        System.out.print(large);
        scanner.close();
    }
}

/*
For example, if we create an hourglass using the number 1 within an array full of zeros, it may look like this:

1 1 1 0 0 0
0 1 0 0 0 0
1 1 1 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0

Actually, there are many hourglasses in the array above. The three leftmost hourglasses are the following:

1 1 1     1 1 0     1 0 0
  1         0         0
1 1 1     1 1 0     1 0 0

Input Format

There will be exactly
lines, each containing integers seperated by spaces. Each integer will be between and

inclusive.

Output Format

Print the answer to this problem on a single line.

Sample Input

1 1 1 0 0 0
0 1 0 0 0 0
1 1 1 0 0 0
0 0 2 4 4 0
0 0 0 2 0 0
0 0 1 2 4 0

Sample Output

19

Explanation

The hourglass which has the largest sum is:

2 4 4
  2
1 2 4
*/