# Java Arraylist HackerRank Solution Hello Programmers, In this post, you will know how to solve the Java Arraylist HackerRank Solution. This problem is a part of the HackerRank Java Programming Series.

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.

## Java Arraylist HackerRank Solution

### Problem

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  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  position of  line.

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

Constraints

Each number will fit in signed integer.
Total number of integers in  lines will not cross .

Output Format
In each line, output the number located in  position of  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
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:

### Java Arraylist HackerRank Solutions

```import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
List<List<Integer>> lines = new ArrayList<List<Integer>>();
int n = in.nextInt();
for (int i = 0; i < n; i++) {
List<Integer> line = new ArrayList<Integer>();
int d = in.nextInt();
for (int j = 0; j < d; j++) {
}
}
int q = in.nextInt();
for (int i = 0; i < q; i++) {
int x = in.nextInt();
int y = in.nextInt();
if (y > lines.get(x - 1).size()) {
System.out.println("ERROR!");
} else {
System.out.println(lines.get(x - 1).get(y - 1));
}
}
in.close();
}
}```

Disclaimer: The above Problem (Java Arraylist) is generated by Hackerrank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.