Problem Description :-
Let's play a game on an array! You're standing at index 0 of an n-element array names 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, i - 1 or i + leap 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.
Given leap and game, complete the function in the editor below so that it returns true if you can win the game (or false if you cannot).
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Lets see solution of problem.
Solution 1 :- Using Stack Data Structure
import java.util.*;
public class Solution {
public static boolean canWin(int leap, int[] game) {
int arrayLength = game.length;
Stack<Integer> stack = new Stack<>();
stack.push(0);
while (!stack.isEmpty()) {
int i = stack.pop();
if (i >= arrayLength) return true;
if (i < 0 || game[i] == 1) continue;
game[i] = 1;
stack.push(i+1);
stack.push(i-1);
stack.push(i+leap);
}
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();
}
}
Output :-
5 3
0 0 0 0 0
YES
-----------
6 3
0 0 1 1 1 0
NO
Explanation :-
- ArrayLength = 6, leap = 3, game = [0 0 1 1 1 0]
- Defined stack data structure. push 0 value into stack for initial.
- While loop until stack is not empty.
- Get first value from stack using pop operation. value = 0.
- i >= arrayLength, 0 >= 6, No.
- i < 0 || game[0] == 1, 0 < 0 || 0 == 1, No,
- game[i] = 1, set 1 to 0th index in game array, [1 0 1 1 1 0].
- Push i + 1 in stack, 0 + 1 = 1.
- Push i - 1 in stack, 0 - 1 = -1.
- Push i + leap in stack, 0 + 3 = 3.
- Stack = [1 -1 3], game = [1 0 1 1 1 0].
- Stack is not empty.
- Pop 3.
- i >= arrayLength, 3 >= 6, No.
- i < 0 || game[3] == 1, 3 < 0 || 1 == 1, Yes, Continue.
- Stack is not empty.
- Pop -1.
- i >= arrayLength, -1 >= 6, No.
- i < 0 || game[-1] == 1, -1 < 0 , Yes, Continue.
- Stack is not empty.
- Pop 1.
- i >= arrayLength, 1 >= 6, No.
- i < 0 || game[1] == 1, 1 < 0 || 0 == 1, No.
- game[i] = 1, set 1 to 0th index in game array, [1 1 1 1 1 0].
- Push i + 1 in stack, 1 + 1 = 2.
- Push i - 1 in stack, 1 - 1 = 0.
- Push i + leap in stack, 1 + 3 = 4.
- Stack = [2 0 4], game = [1 1 1 1 1 0].
- Stack is not empty.
- Pop 4.
- i >= arrayLength, 4 >= 6, No.
- i < 0 || game[1] == 1, 4 < 0 || 1 == 1, Yes, Continue.
- Stack is not empty.
- Pop 0.
- i >= arrayLength, 1 >= 6, No.
- i < 0 || game[-1] == 1, 1 < 0 , 1 == 1 Yes, Continue
- Stack is not empty.
- Pop 2.
- i >= arrayLength, 2 >= 6, No.
- i < 0 || game[-1] == 1, 2 < 0 , 1 == 1 Yes, Continue
- Stack is empty
- Return false and print NO.
Solution 2 :- Using Recursive Approach
import java.util.*;
public class Solution {
public static boolean canWin(int leap, int[] game, int i) {
// Return true if you can win the game; otherwise, return false.
if (i >= game.length) {
return true;
} else if (i < 0 || game[i] == 1) {
return false;
}
game[i] = 1;
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();
}
}
Happy Coding.
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