HackerRank Java Arraylist problem solution

In this HackerRank java Arraylist problem in java programming language 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 the Yth position of the Xth line.

HackerRank Java Arraylist problem solution.

```import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

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 scan=new Scanner(System.in);

int total=scan.nextInt();

ArrayList<ArrayList<Integer>> mainlist=new ArrayList<>();

for(int i=0;i<total;i++)
{
int size=scan.nextInt();
ArrayList<Integer> list=new ArrayList();

for(int j=0;j<size;j++)
{
int item=scan.nextInt();
}
}

int totalprint=scan.nextInt();
for(int k=0;k<totalprint;k++)
{
int row=scan.nextInt();
int col=scan.nextInt();
try
{
System.out.println(mainlist.get(row-1).get(col-1));
}
catch(Exception e)
{
System.out.println("ERROR!");
}
}
scan.close();
}
}```

Second solution

```import java.io.BufferedReader;
import java.io.IOException;
import java.util.List;
import java.util.ArrayList;

public class Solution
{
public static void main(String[] args) throws IOException {

final int N = Integer.parseInt(br.readLine().trim(), 10);

final List<List<Integer>> arr = new ArrayList<List<Integer>>();

for (int i = 0; i < N; i++) {
final String[] data = br.readLine().trim().split(" ");
final int D = Integer.parseInt(data[0], 10);
final List<Integer> a = new ArrayList<Integer>();
for (int j = 1; j <= D; j++) {
final int v = Integer.parseInt(data[j], 10);
}
}

final StringBuilder sb = new StringBuilder();
final int Q = Integer.parseInt(br.readLine().trim(), 10);
for (int i = 0; i < Q; i++) {
final String[] query = br.readLine().trim().split(" ");
final int r = Integer.parseInt(query[0], 10);
final int c = Integer.parseInt(query[1], 10);
if (r > arr.size()) {
sb.append("ERROR!\n");
} else {
final List<Integer> row = arr.get(r - 1);
if (c > row.size()) {
sb.append("ERROR!\n");
} else {
sb.append(row.get(c - 1)).append('\n');
}
}
}

System.out.print(sb.toString());

br.close();
br = null;
}
}```

The solution in java8 programming.

```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. */
ArrayList<ArrayList< Integer >> arr = new ArrayList<ArrayList< Integer >>();

Scanner sc = new Scanner(System.in);

int N = sc.nextInt();
for (int n=0 ; n<N ; ++n) {

int D = sc.nextInt();
ArrayList< Integer > a = new ArrayList< Integer >();

for (int d=0 ; d<D ; ++d) {
}

}

int Q = sc.nextInt();
for (int q=0 ; q<Q ; ++q) {

int x = sc.nextInt();
int y = sc.nextInt();

try {
System.out.println(arr.get(x-1).get(y-1));
} catch (Exception e) {
System.out.println("ERROR!");
}
}
}
}```