Game of stones

Given an integer N which is the number of stones in a pile and the game of stones is being played between you and your friend, the task is to find whether you’ll win or not. A player (on his/her turn) can remove 1 or 2 or 3 coins, the game continues in alternating turns. The one who is not able to make the move loses the game. In other words, the player who removes the last set of coins always wins. Print YES if you can win the game else print NO. 

Examples:

Input: N = 4 
Output: NO 
Explanation: If there are 4 stones in the bag, then you will never win the game. No matter 1, 2, or 3 stones you remove, the last stone will always be removed by your friend. Hence, you are not able to make the move.

Input: N = 3 
Output: YES 
Explanation: Yes you can win the game, you just remove all 3 stones in your turn. 

Approach:

• First consider base cases, i.e for n = [1, 2, 3]. In these cases, you always win as you can pick all stones and your friend cannot make a move.
• If n = 4, you will lose. Because no matter how many you take, you will leave some stones behind for your friend to take and win the game. So in order to win, you have to ensure that you never reach the situation where there are exactly four stones left when your turn comes up.
• Similarly, if there are 5, 6, or 7 stones you can win by taking just enough to leave 4 stones for your friend. But if there are 8 stones on the pile, you will lose because regardless of whether you pick 1, 2 or 3 stones, your friend can pick 3, 2 or 1 stone to ensure that, again, 4 stones will be left for you.
• It is obvious that the same pattern repeats, so you will always lose if n % 4 == 0.

Below is the implementation of the above approach: 

C++

// C++ program of game of stones
#include <bits/stdc++.h>
using namespace std;

// Function that returns true if u win
bool checkWin(int n)
{
    if (n % 4 != 0)
        return true;
    return false;
}

// Driver code
int main()
{
    // n is number of stones
    int n = 4;

    if (checkWin(n))
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
    return 0;
}

Java

// Java program of game of stones
import java.io.*;

class GFG {
    // Function that returns true if u win
static boolean checkWin(int n)
{
    if (n % 4 != 0)
        return true;
    return false;
}

// Driver code
    public static void main (String[] args) {

    // n is number of stones
    int n = 4;

    if (checkWin(n))
            System.out.println("YES");
    else
            System.out.println( "NO");

    }
//This code is contributed by akt_mit
}

python 3

# Python3 program of game of stones

# Function that returns true if u win
def checkWin( n):

    if (n % 4 != 0):
        return True
    return False

# Driver code
if __name__ == "__main__":

    # n is number of stones
    n = 4

    if (checkWin(n)):
        print ( "YES")
    else:
        print ("NO")

C#

//C#  program of game of stones
using System;

public class GFG{
// Function that returns true if u win
static bool checkWin(int n)
{
    if (n % 4 != 0)
        return true;
    return false;
}

// Driver code

    static public void Main (){

    // n is number of stones
    int n = 4;

    if (checkWin(n))
            Console.WriteLine("YES");
    else
            Console.WriteLine( "NO");

    }
//This code is contributed by ajit
}

PHP

<?php
// PHP program of game of stones

// Function that returns
// true if u win
function checkWin($n)
{
    if ($n % 4 != 0)
        return true;
    return false;
}

// Driver code

// n is number of stones
$n = 4;

if (checkWin($n))
    echo "YES","\n" ;
else
    echo "NO","\n" ;

// This code is contributed
// by ANKITRAI1
?>

Javascript

<script>
    // Javascript program of game of stones

    // Function that returns true if u win
    function checkWin(n)
    {
        if (n % 4 != 0)
            return true;
        return false;
    }

    // n is number of stones
    let n = 4;

    if (checkWin(n))
          document.write("YES");
    else
          document.write( "NO");

</script>

Output

NO

Time Complexity: O(1) 
Auxiliary Space: O(1)