Game of Nim with removal of one stone allowed

Difficulty Level : Basic
Last Updated : 14 Apr, 2021

In Game of Nim, two players take turns removing objects from heaps or the pile of stones.
Suppose two players A and B are playing the game. Each is allowed to take only one stone from the pile. The player who picks the last stone of the pile will win the game. Given N the number of stones in the pile, the task is to find the winner, if player A starts the game.
Examples :

Input : N = 3.
Output : Player A

Player A remove stone 1 which is at the top, then Player B remove stone 2 
and finally player A removes the last stone.

Input : N = 15.
Output : Player A

For N = 1, player A will remove the only stone from the pile and wins the game.
For N = 2, player A will remove the first stone and then player B remove the second or the last stone. So player B will win the game.
So, we can observe player A wins when N is odd and player B wins when N is even.
Below is the implementation of this approach:

C++

// C++ program for Game of Nim with removal
// of one stone allowed.
#include<bits/stdc++.h>
using namespace std;
 
// Return true if player A wins,
// return false if player B wins.
bool findWinner(int N)
{
  // Checking the last bit of N.
  return N&1;
}
 
// Driven Program
int main()
{
  int N = 15;
  findWinner(N)? (cout << "Player A";):
                 (cout << "Player B";);
  return 0;
}

Java

// JAVA Code For Game of Nim with
// removal of one stone allowed
import java.util.*;
 
class GFG {
     
    // Return true if player A wins,
    // return false if player B wins.
    static int findWinner(int N)
    {
      // Checking the last bit of N.
      return N & 1;
    }
     
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int N = 15;
        if(findWinner(N)==1)
            System.out.println("Player A");
        else
             System.out.println("Player B");
           
    }
}
 
// This code is contributed by Arnav Kr. Mandal.

Python3

# Python3 code for Game of Nim with
# removal of one stone allowed.
 
# Return true if player A wins,
# return false if player B wins.
def findWinner( N ):
     
    # Checking the last bit of N.
    return N & 1
     
# Driven Program
N = 15
print("Player A" if findWinner(N) else "Player B")
 
# This code is contributed by "Sharad_Bhardwaj".

C#

// C# Code For Game of Nim with
// removal of one stone allowed
using System;
 
class GFG {
     
    // Return true if player A wins,
    // return false if player B wins.
    static int findWinner(int N)
    {
        // Checking the last bit of N.
        return N & 1;
    }
     
    /* Driver program to test above function */
    public static void Main()
    {
        int N = 15;
         
        if(findWinner(N) == 1)
            Console.Write("Player A");
        else
            Console.Write("Player B");
         
    }
}
 
// This code is contributed by vt_m.

PHP

<?php
// PHP program for Game of
// Nim with removal of one
// stone allowed.
 
// Return true if player A wins,
// return false if player B wins.
function findWinner($N)
{
     
// Checking the last bit of N.
return $N&1;
}
 
// Driver Code
$N = 15;
 
if(findWinner($N))
echo "Player A";
else
echo "Player B";
 
// This code is contributed by vt_m.
?>

Javascript

<script>
 
// JavaScript program For Game of Nim with
// removal of one stone allowed
 
   // Return true if player A wins,
    // return false if player B wins.
    function findWinner(N)
    {
     
      // Checking the last bit of N.
      return N & 1;
    }
 
// Driver code
        let N = 15;
        if(findWinner(N)==1)
            document.write("Player A");
        else
             document.write("Player B");
          
         // This code is contributed by sanjoy_2.
</script>

Output :

Player A

Time Complexity: O(1).
This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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