Interview Preparation Kit - Array(2D Array-DS) hacker rank
2D Array - DS
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. Calculate the hourglass sum for every hourglass in , then print the maximum hourglass sum.
For example, given the 2D array:
-9 -9 -9 1 1 1
0 -9 0 4 3 2
-9 -9 -9 1 2 3
0 0 8 6 6 0
0 0 0 -2 0 0
0 0 1 2 4 0
We calculate the following hourglass values:
-63, -34, -9, 12,
-10, 0, 28, 23,
-27, -11, -2, 10,
9, 17, 25, 18
Our highest hourglass value is from the hourglass:
0 4 3
1
8 6 6
Note: If you have already solved the Java domain's Java 2D Array challenge, you may wish to skip this challenge.
Function Description
Complete the function hourglassSum in the editor below. It should return an integer, the maximum hourglass sum in the array.
hourglassSum has the following parameter(s):
- arr: an array of integers
Input Format
Each of the lines of inputs contains space-separated integers .
Output Format
Print the largest (maximum) hourglass sum found in arr .
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:
The hourglass with the maximum sum () is:
2 4 4
2
1 2 4
And here is java8 solution for the question :
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 main(String args []){Scanner sn = new Scanner(System.in);int a[][] = new int[6][6];for(int i=0;i<6;i++){for(int j=0;j<6;j++){a[i][j] = sn.nextInt();}}int m=0;int b[] = new int[16];for(int i=0;i<4;i++){for(int j=0;j<4;j++){b[m] = a[i][j]+a[i][j+1]+a[i][j+2]+a[i+1][j+1]+a[i+2][j]+a[i+2][j+1]+a[i+2][j+2];m++;}}int max =0;max = b[0];for(int i = 0; i < 16; i++){if(max < b[i]){max = b[i];}}System.out.println(max);}}
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