Predict the winner in Coin Game

There are two players P1 and P2 and two piles of coins consisting of M and N coins respectively. At each turn, a player can choose only one of the piles out of these and discard the other one. This discarded pile cannot be used further in the game. The pile player chooses is further divided into two piles of non-zero parts. The player who cannot divide the pile i.e. the number of coins in the pile is < 2, loses the game. The task is to determine which player wins if P1 starts the game and both the players play optimally.

Examples:

Input: M = 4, N = 4 
Output: Player 1 
Explanation: Player 1 can choose any one of the piles as both contain the same number of coins 
and then splits the chosen one (the one which is not chosen is discarded) into two piles with 1 coin each. 
Now, player 2 is left with no move (as both the remaining piles contain a single coin each 
which cannot be split into two groups of non-zero coins).

Input: M = 1, N = 1 
Output: Player 2 

Approach: Simply check if any of the piles consist of an even number of coins. If yes then Player 1 wins else Player 2 wins.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to print the winner of the game
void findWinner(int M, int N)
{
    if (M % 2 == 0 || N % 2 == 0)
        cout << "Player 1";
    else
        cout << "Player 2";
}

// Driver code
int main()
{
    int M = 1, N = 2;
    findWinner(M, N);

    return 0;
}

Java

// Java implementation of the approach
import java.io.*;

class GFG {

    // Function to print the winner of the game
    static void findWinner(int M, int N)
    {
        if (M % 2 == 0 || N % 2 == 0)
            System.out.println("Player 1");
        else
            System.out.println("Player 2");
    }

    // Driver code
    public static void main(String[] args)
    {
        int M = 1, N = 2;
        findWinner(M, N);
    }
}

// This code is contributed by ajit.

Python3

# Python implementation of the approach
# Function to print the winner of the game

def findWinner(M, N):
    if (M % 2 == 0 or N % 2 == 0):
        print("Player 1");
    else:
        print("Player 2");

# Driver code
M = 1;
N = 2;
findWinner(M, N);


# This code contributed by PrinciRaj1992

C#

// C# implementation of the approach
using System;

class GFG {

    // Function to print the winner of the game
    static void findWinner(int M, int N)
    {
        if (M % 2 == 0 || N % 2 == 0)
            Console.WriteLine("Player 1");
        else
            Console.WriteLine("Player 2");
    }

    // Driver code
    static public void Main()
    {
        int M = 1, N = 2;
        findWinner(M, N);
    }
}

// This code is contributed by Tushil..

PHP

<?php
//PHP implementation of the approach
// Function to print the winner of the game

function findWinner($M, $N)
{
    if ($M % 2 == 0 || $N % 2 == 0)
        echo "Player 1";
    else
        echo "Player 2";
}

    // Driver code
    $M = 1;
    $N = 2;
    findWinner($M, $N);

// This code is contributed by Tushil.
?>

Javascript

<script>

// Javascript implementation of the approach

// Function to print the winner of the game
function findWinner(M, N)
{
    if (M % 2 == 0 || N % 2 == 0)
        document.write(("Player 1"));
    else
        document.write(("Player 2"));
}

// Driver code
var M = 1, N = 2;
findWinner(M, N);

// This code is contributed by rutvik_56.
</script>

Output

Player 1

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