Hackerrank Day 11: 2D Arrays | Hackerrank Solutions in Java

Objective
Today, we're building on our knowledge of Arrays by adding another dimension. Check out the Tutorial tab for learning materials and an instructional video!

Context
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.

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

Explanation

 contains the following hourglasses:

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 hourglass with the maximum sum () is:

2 4 4
  2
1 2 4

Solution:

import java.util.Scanner;

public class Solution {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int arr[][] = new int[6][6];
        for (int i = 0; i < 6; i++) {
            for (int j = 0; j < 6; j++) {
                arr[i][j] = in.nextInt();
            }
        }

        int max = -9 * 7;
        for (int i = 0; i < 6; i++) {
            for (int j = 0; j < 6; j++) {
                if (j + 2 < 6 && i + 2 < 6) {
                    int 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 > max) max = sum;
                }
            }
        }

        System.out.println(max);
    }
}