Hello coders, today we are going to solve Java 2D Array HackerRank Solution.

Problem
You are given a 6 * 6 2D array. An hourglass in an array is a portion shaped like this:
a b c
d
e f g
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
The sum of an hourglass is the sum of all the numbers within it. The sum for the hourglasses above are 7, 4, and 2, respectively.
In this problem you have to print the largest sum among all the hourglasses in the array.
Input Format
There will be exactly 6 lines, each containing 6 integers seperated by spaces. Each integer will be between -9 and 9 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 Input
19
Explanation
The hourglass which has the largest sum is:
2 4 4
2
1 2 4
Solution – Java 2D Array
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 csum =Integer.MIN_VALUE; for(int i=0;i<6;i++) { for(int j=0;j<6;j++) { if(i>1 && j>1) { int sum=arr[i][j]+arr[i-1][j-1]+arr[i-2][j]+arr[i-2][j-1] +arr[i-2][j-2]+arr[i][j-1]+arr[i][j-2]; if(sum>csum) { csum=sum; } } } } System.out.print(csum); scanner.close(); } }
Disclaimer: The above Problem ( Java 2D Array ) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.