Java 1D Array (Part 2)

Let’s play a game on an array! You’re standing at index 0 of an n-element array named game. From some index i (where 0≤i<n), you can perform one of the following moves:

• Move Backward: If cell i-1 exists and contains a 0, you can walk back to cell i-1.
• Move Forward:
  • If cell i+1 contains a zero, you can walk to cell i+1.
  • If cell i+leap contains a zero, you can jump to cell  i+leap .
  • If you’re standing in cell n-1 or the value of  i+leap≥n , you can walk or jump off the end of the array and win the game.

In other words, you can move from index i to index , i+1, or  as long as the destination index is a cell containing a 0. If the destination index is greater than n-1, you win the game.

Function Description

Complete the canWin function in the editor below.

canWin has the following parameters:

• int leap: the size of the leap
• int game[n]: the array to traverse

Returns

• boolean: true if the game can be won, otherwise false

Input Format

The first line contains an integer, q, denoting the number of queries (i.e., function calls).
The 2.q subsequent lines describe each query over two lines:

1. The first line contains two space-separated integers describing the respective values of n and leap.
2. The second line contains  space-separated binary integers (i.e., zeroes and ones) describing the respective values of game, game1,…, gamen-1.

Constraints

• 1 ≤ q≤5000
• 2≤ n ≤100
• 0 ≤ leap < 100
• It is guaranteed that the value of game[0] is always 0.

Sample Input

STDIN           Function
-----           --------
4               q = 4 (number of queries)
5 3             game[] size n = 5, leap = 3 (first query)
0 0 0 0 0       game = [0, 0, 0, 0, 0]
6 5             game[] size n = 6, leap = 5 (second query)
0 0 0 1 1 1     . . .
6 3
0 0 1 1 1 0
3 1
0 1 0

Sample Output

YES
YES
NO
NO

Explanation

We perform the following q=4 queries:

1. For game = [0,0,0,0,0] and leap = 3, we can walk and/or jump to the end of the array because every cell contains a 0. Because we can win, we return true.
2. For game = [0,0,0,1,1,1] and leap = 5, we can walk to index 1 and then jump i+leap = 1+5 =6 units to the end of the array. Because we can win, we return true.
3. For game = [0,0,1,1,1,0] and leap =3, there is no way for us to get past the three consecutive ones. Because we cannot win, we return false.
4. For game = [0,1,0] and leap = 1, there is no way for us to get past the one at index 1. Because we cannot win, we return false.

Solution Implementation

import java.util.*;
public class Solution { 
  public static boolean canWin(int leap, int[] game) {
    Queue<Integer> stack = new LinkedList<>();
    stack.add(0);

    while (!stack.isEmpty()) {
      int current = stack.poll();
      game[current] = 2;
      if (current + 1 >= game.length || current + leap >= game.length) {
        return true;
      }

      if (game[current + 1] == 0) {
        stack.add(current + 1);
      }

      if (game[current + leap] == 0) {
        stack.add(current + leap);
      }

      if (current - 1 >= 0 && game[current - 1] == 0) {
        stack.add(current - 1);
      }
    }

    return false;
  }

  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)) ? "YES" : "NO" );
    }
    scan.close();
  }
}