Java ArrayList | HackerRank Solution

Hello coders, today we are going to solve Java ArrayList HackerRank Solution.

Java ArrayList

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 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 105.

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 Input

 74
 52
 37
 ERROR!
 ERROR!

Explanation

The diagram below explains the queries:

Java ArrayList Explanation

Solution – Java ArrayList

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

public class Solution {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        String s = sc.nextLine();
        ArrayList<ArrayList<String>> a = new ArrayList<ArrayList<String>>(n);
        for (int i = 0; i < n; i++) {
            s = sc.nextLine();
            a.add(i, new ArrayList<String>(Arrays.asList(s.split("\\s"))));
        }

        int m = sc.nextInt();
        for (int i = 0; i < m; i++) {
            int x = sc.nextInt();
            int y = sc.nextInt();
            if (x <= a.size() && y < a.get(x-1).size() && y >= 0) {
                System.out.println(a.get(x-1).get(y));
            } else {
                System.out.println("ERROR!");
            }
        }
    }
}

Disclaimer: The above Problem ( Java ArrayList ) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.

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