Maximum sum of hour glass in matrix

Problem:

Given a 6 x 6 2D array, arr.

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

An hourglass in A is a subset of values with indices facing in this pattern in arr‘s graphical representation.

a b c
  d
e f g

There are 16 hourglass in arr. An hourglass is the sum of an hourglass values. Calculate the hourglass sum for every hourglass in arr, then print maximum hourglass sum. The array always be 6x6.

For example, for the below 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

The 16 hourglass sums are

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

The highest hourglass sum is 28 from the hourglass beginning at row 1, column 2.

0 4 3
  1
8 6 6 

Function Description:

Complete the function hourglassSum which has the following parameters.

int arr[6][6] : an array of intergers

Returns

int: the maximum hourglass sum

Input Format

Each of the 6 lines of inputs arr[i] contains 6 space separated integers arr[i][j]

Constraints

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

Output Format:

Print the largest (maximum) hourglass sum found in arr.

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

Output

19

Explanation:

All hour glasses are as below

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

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

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

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

The hour glass with maximum sum is below, and the sum is 19.

2 4 4
  2
1 2 4

Solution

package org.span;

import java.io.IOException;
import java.util.Scanner;

public class HourGlass {

    static int hourglassSum(int[][] arr) {
        int max = -99999, cur;
        for (int i = 0; i < 4; i++) {
            for (int j = 0; j < 4; j++) {
                cur = 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 (cur > max) {
                    max = cur;
                }
            }
        }
        return max;
    }

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

    public static void main(String[] args) throws IOException {
        /*int[][] arr = {
                {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}
        };*/
        int[][] arr = {
                {0, -4, -6, 0, -7, -6},
                {-1, -2, -6, -8, -3, -1},
                {-8, -4, -2, -8, -8, -6},
                {-3, -1, -2, -5, -7, -4},
                {-3, -5, -3, -6, -6, -6},
                {-3, -6, 0, -8, -6, -7},
        };
        int result = hourglassSum(arr);
        System.out.println(result);
    }
}