# Game of Nim with removal of one stone allowed

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 write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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