# Java 1D Array (Part 2) HackerRank Solution

#### ByBrokenprogrammers

Dec 4, 2022

Hello Programmers, In this post, you will know how to solve the Java 1D Array (Part 2) HackerRank Solution. This problem is a part of the HackerRank Java Programming Series.

One more thing to add, donâ€™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 1D Array (Part 2) HackerRank Solutions

### Problem

Let’s play a game on an array! You’re standing at index  of an -element array named . From some index  (where ), you can perform one of the following moves:

• Move Backward: If cell  exists and contains a , you can walk back to cell .

### Java 1D Array (Part 2) HackerRank Solutions

```import java.util.*;
public class Solution {
public static boolean canWin(int leap, int[] game, int i) {
if (i < 0 || game[i] == 1)
return false;
if (i + 1 >= game.length || i + leap >= game.length)
return true;
game[i] = 1; //flag
return canWin(leap, game, i + leap)
|| canWin(leap, game, i + 1)
|| canWin(leap, game, i - 1);
}
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int q = scan.nextInt();
while (q-- > 0) {
int n = scan.nextInt();
int leap = scan.nextInt();
int[] game = new int[n];
for (int i = 0; i < n; i++) {
game[i] = scan.nextInt();
}
System.out.println((canWin(leap, game, 0)) ? "YES" : "NO");
}
scan.close();
}
}```

Disclaimer: The above Problem (Java 1D Array (Part 2)) is generated by Hackerrank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.