Maximize big when both big and small can be exchanged

Given N Big Candies and M Small Candies. One Big Candy can be bought by paying X small candies. Alternatively, one big candy can be sold for Y small candies. The task is to find the maximum number of big candies that can be bought.
Examples:

Input: N = 3, M = 10, X = 4, Y = 2 
Output: 5 
8 small candies are exchanged for 2 big candies.
Input: N = 3, M = 10, X = 1, Y = 2 
Output: 16 
Sell all the initial big candies to get 6 small candies. 
Now 16 small candies can be exchanged for 16 big candies. 

In first example, Big candies cannot be sold for profit. So, only the remaining small candies can be exchanged for big candies. 
In second example, Big candies can be sold for profit.
Approach: If initial big candies can be sold for profit i.e. X < Y then sell the big candies and update the count of small and big candies. Then, sell all of the updated small candies in order to buy big candies.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <iostream>
using namespace std;

    // Function to return the maximum big
    // candies that can be bought
    int max_candies(int bigCandies,
        int smallCandies,int X, int Y)
    {
        // If initial big candies
        // can be sold for profit
        if (X < Y)
        {
            smallCandies += Y * bigCandies;
            bigCandies = 0;
        }

        // Update big candies that can be bought
        bigCandies += (smallCandies / X);

        return bigCandies;
    }

    // Driver code
    int main()
    {
        int N = 3, M = 10;
        int X = 4, Y = 2;
        cout << (max_candies(N, M, X, Y));
        return 0;
    }

Java

// Java implementation of the approach
class GFG {

    // Function to return the maximum big candies
    // that can be bought
    static int max_candies(int bigCandies, int smallCandies,
                           int X, int Y)
    {
        // If initial big candies can be sold for profit
        if (X < Y) {

            smallCandies += Y * bigCandies;
            bigCandies = 0;
        }

        // Update big candies that can be bought
        bigCandies += (smallCandies / X);

        return bigCandies;
    }

    // Driver code
    public static void main(String[] args)
    {
        int N = 3, M = 10;
        int X = 4, Y = 2;

        System.out.println(max_candies(N, M, X, Y));
    }
}

Python3

# Python3 implementation of the approach

# Function to return the maximum big candies
# that can be bought
def max_candies(bigCandies, smallCandies, X, Y):

    # If initial big candies can
    # be sold for profit
    if(X < Y):

        smallCandies += Y * bigCandies
        bigCandies = 0

    # Update big candies that can be bought
    bigCandies += (smallCandies // X)

    return bigCandies

# Driver code
N = 3
M = 10
X = 4
Y = 2
print(max_candies(N, M, X, Y))

# This code is contributed by Code_Mech

C#

// C# implementation of the approach
using System;

class GFG
{

    // Function to return the maximum
    // big candies that can be bought
    static int max_candies(int bigCandies,
                        int smallCandies,
                        int X, int Y)
    {
        // If initial big candies
        // can be sold for profit
        if (X < Y)
        {
            smallCandies += Y * bigCandies;
            bigCandies = 0;
        }

        // Update big candies that can be bought
        bigCandies += (smallCandies / X);

        return bigCandies;
    }

    // Driver code
    static public void Main ()
    {
        int N = 3, M = 10;
        int X = 4, Y = 2;
        Console.WriteLine(max_candies(N, M, X, Y));
    }
}

// This Code is contributed by ajit...

PHP

<?php
// PHP implementation of the approach

// Function to return the maximum big
// candies that can be bought
function max_candies($bigCandies,
                     $smallCandies, $X, $Y)
{
    // If initial big candies can be
    // sold for profit
    if ($X < $Y)
    {

        $smallCandies += $Y * $bigCandies;
        $bigCandies = 0;
    }

    // Update big candies that can be bought
    $bigCandies += (int)($smallCandies / $X);

    return $bigCandies;
}

// Driver code
$N = 3;
$M = 10;
$X = 4;
$Y = 2;

echo (max_candies($N, $M, $X, $Y));

// This code is contributed by akt_mit
?>

Javascript

<script>
    // Javascript implementation of the approach

    // Function to return the maximum
    // big candies that can be bought
    function max_candies(bigCandies, smallCandies, X, Y)
    {
        // If initial big candies
        // can be sold for profit
        if (X < Y)
        {
            smallCandies += Y * bigCandies;
            bigCandies = 0;
        }

        // Update big candies that can be bought
        bigCandies += parseInt(smallCandies / X, 10);

        return bigCandies;
    }

    let N = 3, M = 10;
    let X = 4, Y = 2;
    document.write(max_candies(N, M, X, Y));

</script>

Output:

5

Time Complexity: O(1) as constant operations done.

Space Complexity: O(1) as no extra space has been used.