2D Array - DS

Hourglass.java

/**
 Given a  2D Array, :

 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
 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.

 Task 
 Calculate the hourglass sum for every hourglass in , then print the maximum hourglass sum.

 Note: If you have already solved the Java domain's Java 2D Array challenge, you may wish to skip this challenge.

 Input Format

 There are  lines of input, where each line contains  space-separated integers describing 2D Array ; every value in  will be in the inclusive range of  to .

 Constraints

 Output Format

 Print the largest (maximum) hourglass sum found in .

 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
 */
public class Hourglass {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int arr[][] = new int[6][6];
        for (int arr_i = 0; arr_i < 6; arr_i++) {
            for (int arr_j = 0; arr_j < 6; arr_j++) {
                arr[arr_i][arr_j] = in.nextInt();
            }
        }
        System.out.print(getMax(arr));
    }

    private static int getMax(int[][] arr) {
        int max = 0;
        for (int i = 0; i < 4; i++) {
            for (int j = 0; j < 4; j++) {
                int local_max = 0;
                local_max += arr[i][j];
                local_max += arr[i + 1][j];
                local_max += arr[i + 2][j];
                local_max += arr[i + 1][j + 1];
                local_max += arr[i][j + 2];
                local_max += arr[i + 1][j + 2];
                local_max += arr[i + 2][j + 2];
                if (local_max > max) {
                    max = local_max;
                }
            }
        }
        return max;
    }
}