Java Arraylist HackerRank Solution

Java Arraylist HackerRank Solution
Java Arraylist HackerRank Solution

Java Arraylist HackerRank Solution

Sometimes it’s better to use dynamic size arrays. Java’s Arraylist can provide you this feature. Try to solve this problem using Arraylist.

You are given n lines. In each line there are zero or more integers. You need to answer a few queries where you need to tell the number located in yth position of xth line.

Take your input from System.in.

Input Format
The first line has an integer n. In each of the next n lines there will be an integer d denoting number of integers on that line and then there will be d space-separated integers. In the next line there will be an integer q denoting number of queries. Each query will consist of two integers x and y.

Constraints

  • 1<=n<=20000
  • 0<=d<=50000
  • 1<=q<=1000
  • 1<=x<=n

Each number will fit in signed integer.
Total number of integers in n lines will not cross 10^5.

Output Format
In each line, output the number located in yth position of xth line. If there is no such position, just print “ERROR!”

Sample Input

5
5 41 77 74 22 44
1 12
4 37 34 36 52
0
3 20 22 33
5
1 3
3 4
3 1
4 3
5 5

Sample Output

74
52
37
ERROR!
ERROR!

Explanation

The diagram below explains the queries:

image
Image Credit Hackerrank.com

Code Solution:

#Java Arraylist HackerRank Solution

import java.io.*;
import java.util.*;

public class Solution {

    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
        Scanner in = new Scanner(System.in);
    int n = in.nextInt();
    int d,q,x,y;
    ArrayList[] set = new ArrayList[n];
    for(int i=0;i<n;i++){
        d = in.nextInt();
        set[i] = new ArrayList();
        for(int j=0;j<d;j++){  
            set[i].add(in.nextInt());                
        }
    }
    q=in.nextInt();
    for(int i=0;i<q;i++){
        x=in.nextInt();
        y=in.nextInt();
        try{
            System.out.println(set[x-1].get(y-1));
        } catch(Exception e){
            System.out.println("ERROR!");
        }
    }
    }
}


#Java Arraylist HackerRank Solution

Disclaimer: This problem is originally created and published by HackerRank, we only provide solutions to this problem. Hence, doesn’t guarantee the truthfulness of the problem. This is only for information purposes.

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