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Link for the Problem – Day 11: 2D Arrays – Hacker Rank Solution
Day 11: 2D Arrays – Hacker Rank Solution
Problem:
Objective
Today, we are 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.
Example
In the array shown above, the maximum hourglass sum is for the hourglass in the top left corner.
Input Format
There are lines of input, where each line contains space-separated integers that describe the 2D Array .
Constraints
Output Format
Print the maximum hourglass sum 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
Day 11: 2D Arrays – Hacker Rank Solution
import java.util.Scanner;
/**
* @author Techno-RJ
*
*/
public class Day112DArrays {
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();
}
}
in.close();
int sum = 0, maxSum = Integer.MIN_VALUE;
for (int i = 0; i < 6; i++) {
for (int j = 0; j < 6; j++) {
if ((i + 2) < 6 && (j + 2) < 6) {
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 > maxSum) {
maxSum = sum;
}
}
}
System.out.println(maxSum);
}
}
