Find the player who will win the Coin game

Given N coins, the task is to find who win the coin game.
Coin game is a game in which each player picks coins from the given N coins in such a way that he can pick coins ranging from 1 to 5 coins in one turn and the game continues for both the players. The player who picks the last coin loses the game.
Examples:

Input: N = 4
Output: First Player
Explanation:
Player 1 pick 3 coins and 
Player 2 pick last coin

Input: N = 7
Output: Second Player

Approach:

• As the player can take coins ranging from 1 to 5 inclusively and if a player loses it means that he had only 1 coin, otherwise, he could have taken 1 less coin than available coins and force another player to lose. So now we will consider the case when the second player is going to win, which means the first player had only one coin.
• For N = 1, second player is going to win, for N = 2 to 6, first player can choose 1 less coin than N, and force the second player to lose so discard them, for N = 7, first player can choose coin 1 to 5, that’s going to leave coin ranging from 6 to 2, and then second player can choose 1 to 5, and to win second player will intelligently choose 1 less coin forcing first to lose, So basically starting from 1, all the numbers on a gap of 6(as whatever first player choose, second player will choose coins equal to difference of 6 and coins chosen by the first player) will be the winning for second player.
• Finally, we just have to check if n is of form 6*c+1 if it is then the second player going to win, otherwise, the first player going to win.

Below is the implementation for the above approach: 

C++

// C++ program to find the player
// who wins the game

#include <bits/stdc++.h>
using namespace std;

// Function to check the
// winning player
void findWinner(int n)
{
    // As discussed in the
    // above approach
    if ((n - 1) % 6 == 0) {
        cout << "Second Player wins the game";
    }
    else {
        cout << "First Player wins the game";
    }
}

// Driver function
int main()
{

    int n = 7;
    findWinner(n);
}

Java

// Java program to find the player
// who wins the game
class GFG
{

// Function to check the
// winning player
static void findWinner(int n)
{
    // As discussed in the
    // above approach
    if ((n - 1) % 6 == 0)
    {
        System.out.println("Second Player wins the game");
    }
    else
    {
        System.out.println("First Player wins the game");
    }
}

// Driver Code
public static void main(String[] args)
{
    int n = 7;
    findWinner(n);
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 program to find the player
# who wins the game

# Function to check the
# winning player
def findWinner(n):

    # As discussed in the
    # above approach
    if ((n - 1) % 6 == 0):
        print("Second Player wins the game");
    else:
        print("First Player wins the game");

# Driver Code
if __name__ == '__main__':
    n = 7;
    findWinner(n);

# This code is contributed by 29AjayKumar

C#

// C# program to find the player
// who wins the game

using System;

class GFG
{

    // Function to check the
    // winning player
    static void findWinner(int n)
    {
        // As discussed in the
        // above approach
        if ((n - 1) % 6 == 0)
        {
            Console.WriteLine("Second Player wins the game");
        }
        else
        {
            Console.WriteLine("First Player wins the game");
        }
    }

    // Driver Code
    public static void Main()
    {
        int n = 7;
        findWinner(n);
    }
}

// This code is contributed by AnkitRai01

Javascript

<script>

// JavaScript program to find the player
// who wins the game

// Function to check the
// winning player
function findWinner(n)
{
    // As discussed in the
    // above approach
    if ((n - 1) % 6 == 0) {
        document.write("Second Player wins the game");
    }
    else {
        document.write("First Player wins the game");
    }
}

// Driver function
var n = 7;
findWinner(n);

</script>

Output: 

Second Player wins the game

Time Complexity:

Auxiliary Space: O(1)