Coin game winner where every player has three choices

A and B are playing a game. In the beginning, there are n coins. Given two more numbers x and y. In each move, a player can pick x or y or 1 coin. A always starts the game. The player who picks the last coin wins the game or the person who is not able to pick any coin loses the game. For a given value of n, find whether A will win the game or not if both are playing optimally.

Examples:

Input :  n = 5, x = 3, y = 4
Output : A
There are 5 coins, every player can pick 1 or
3 or 4 coins on his/her turn.
A can win by picking 3 coins in first chance.
Now 2 coins will be left so B will pick one 
coin and now A can win by picking the last coin.

Input : 2 3 4
Output : B

Let us take few example values of n for x = 3, y = 4.
n = 0 A can not pick any coin so he losses
n = 1 A can pick 1 coin and win the game
n = 2 A can pick only 1 coin. Now B will pick 1 coin and win the game
n = 3 4 A will win the game by picking 3 or 4 coins
n = 5, 6 A will choose 3 or 4 coins. Now B will have to choose from 2 coins so A will win.
We can observe that A wins game for n coins only when B loses for coins n-1 or n-x or n-y.

Try It Yourself

C++

// C++ program to find winner of game
// if player can pick 1, x, y coins
#include <bits/stdc++.h>
using namespace std;

// To find winner of game
bool findWinner(int x, int y, int n)
{
    // To store results
    int dp[n + 1];

    // Initial values
    dp[0] = false;
    dp[1] = true;

    // Computing other values.
    for (int i = 2; i <= n; i++) {

        // If A losses any of i-1 or i-x
        // or i-y game then he will
        // definitely win game i
        if (i - 1 >= 0 and !dp[i - 1])
            dp[i] = true;
        else if (i - x >= 0 and !dp[i - x])
            dp[i] = true;
        else if (i - y >= 0 and !dp[i - y])
            dp[i] = true;

        // Else A loses game.
        else
            dp[i] = false;
    }

    // If dp[n] is true then A will
    // game otherwise  he losses
    return dp[n];
}

// Driver program to test findWinner();
int main()
{
    int x = 3, y = 4, n = 5;
    if (findWinner(x, y, n))
        cout << 'A';
    else
        cout << 'B';

    return 0;
}

Java

// Java program to find winner of game
// if player can pick 1, x, y coins
import java.util.Arrays;

public class GFG { 
    
    // To find winner of game
    static boolean findWinner(int x, int y, int n)
    {
        // To store results
        boolean[] dp = new boolean[n + 1];
     
        Arrays.fill(dp, false);
    
        // Initial values
        dp[0] = false;
        dp[1] = true;
     
        // Computing other values.
        for (int i = 2; i <= n; i++) {
     
            // If A losses any of i-1 or i-x
            // or i-y game then he will
            // definitely win game i
            if (i - 1 >= 0 && dp[i - 1] == false)
                dp[i] = true;
            else if (i - x >= 0 && dp[i - x] == false)
                dp[i] = true;
            else if (i - y >= 0 && dp[i - y] == false)
                dp[i] = true;
     
            // Else A loses game.
            else
                dp[i] = false;
        }
     
        // If dp[n] is true then A will
        // game otherwise  he losses
        return dp[n];
    }
     
    // Driver program to test findWinner();
    public static void main(String args[])
    {
        int x = 3, y = 4, n = 5;
        if (findWinner(x, y, n) == true)
            System.out.println('A');
        else
            System.out.println('B');
    }
}
// This code is contributed by Sumit Ghosh

Python3

# Python3 program to find winner of game
# if player can pick 1, x, y coins

# To find winner of game
def findWinner(x, y, n):
    
    # To store results
    dp = [0 for i in range(n + 1)]

    # Initial values
    dp[0] = False
    dp[1] = True

    # Computing other values.
    for i in range(2, n + 1):

        # If A losses any of i-1 or i-x
        # or i-y game then he will
        # definitely win game i
        if (i - 1 >= 0 and not dp[i - 1]):
            dp[i] = True
        elif (i - x >= 0 and not dp[i - x]):
            dp[i] = True
        elif (i - y >= 0 and not dp[i - y]):
            dp[i] = True

        # Else A loses game.
        else:
            dp[i] = False

    # If dp[n] is true then A will
    # game otherwise he losses
    return dp[n]

# Driver Code
x = 3; y = 4; n = 5
if (findWinner(x, y, n)):
    print('A')
else:
    print('B')

# This code is contributed by Azkia Anam

// C# program to find winner of game
// if player can pick 1, x, y coins
using System;

public class GFG { 
    
    // To find winner of game
    static bool findWinner(int x, int y, int n)
    {
        
        // To store results
        bool[] dp = new bool[n + 1];
    
        for(int i = 0; i < n+1; i++)
            dp[i] =false;
    
        // Initial values
        dp[0] = false;
        dp[1] = true;
    
        // Computing other values.
        for (int i = 2; i <= n; i++)
        {
    
            // If A losses any of i-1 or i-x
            // or i-y game then he will
            // definitely win game i
            if (i - 1 >= 0 && dp[i - 1] == false)
                dp[i] = true;
            else if (i - x >= 0 && dp[i - x] == false)
                dp[i] = true;
            else if (i - y >= 0 && dp[i - y] == false)
                dp[i] = true;
    
            // Else A loses game.
            else
                dp[i] = false;
        }
    
        // If dp[n] is true then A will
        // game otherwise he losses
        return dp[n];
    }
    
    // Driver program to test findWinner();
    public static void Main()
    {
        int x = 3, y = 4, n = 5;
        
        if (findWinner(x, y, n) == true)
            Console.WriteLine('A');
        else
            Console.WriteLine('B');
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP program to find winner of game
// if player can pick 1, x, y coins

// To find winner of game
function findWinner( $x, $y, $n)
{
    
    // To store results
    $dp= array();

    // Initial values
    $dp[0] = false;
    $dp[1] = true;

    // Computing other values.
    for ($i = 2; $i <= $n; $i++)
    {

        // If A losses any of i-1 or i-x
        // or i-y game then he will
        // definitely win game i
        if ($i - 1 >= 0 and !$dp[$i - 1])
            $dp[$i] = true;
        else if ($i - $x >= 0 and !$dp[$i - $x])
            $dp[$i] = true;
        else if ($i - $y >= 0 and !$dp[$i - $y])
            $dp[$i] = true;

        // Else A loses game.
        else
            $dp[$i] = false;
    }

    // If dp[n] is true then A will
    // game otherwise he losses
    return $dp[$n];
}

// Driver program to test findWinner();
    $x = 3; $y = 4; $n = 5;
    if (findWinner($x, $y, $n))
        echo 'A';
    else
        echo 'B';
        
// This code is contributed by anuj_67.
?>

JavaScript

<script>

// Javascript program to find winner of game
// if player can pick 1, x, y coins

// To find winner of game
function findWinner(x, y, n)
{
    
    // To store results
    var dp = Array(n + 1).fill(0);

    // Initial values
    dp[0] = false;
    dp[1] = true;

    // Computing other values.
    for(var i = 2; i <= n; i++) 
    {
        
        // If A losses any of i-1 or i-x
        // or i-y game then he will
        // definitely win game i
        if (i - 1 >= 0 && !dp[i - 1])
            dp[i] = true;
        else if (i - x >= 0 && !dp[i - x])
            dp[i] = true;
        else if (i - y >= 0 && !dp[i - y])
            dp[i] = true;

        // Else A loses game.
        else
            dp[i] = false;
    }

    // If dp[n] is true then A will
    // game otherwise he losses
    return dp[n];
}

// Driver code
var x = 3, y = 4, n = 5;
if (findWinner(x, y, n))
    document.write('A');
else
    document.write('B');

// This code is contributed by noob2000

</script>

Output

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

Coin game winner where every player has three choices

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