Day11-2D-Arrays

Problem:

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 6x6 2D Array,A :

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 A to be a subset of values with indices falling in this pattern in A's graphical representation:

a b c
  d
e f g
There are 16 hourglasses in A , and an hourglass sum is the sum of an hourglass' values.

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

Solution:

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 {
    public static void countMaxSum(int[][] arr) {
        int max_sum = Integer.MIN_VALUE;
        for (int i = 0; i < arr.length - 2; i++) {
            for (int j = 0; j < arr[i].length - 2; j++) {
                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]);
                max_sum = Math.max(max_sum, sum);
            }
        }
        System.out.println(max_sum);

    }


    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;
            }
        }
        Solution.countMaxSum(arr);
        scanner.close();
    }
}