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 approachclass 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 boughtdef 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 codeN = 3M = 10X = 4Y = 2print(max_candies(N, M, X, Y))# This code is contributed by Code_Mech |
C#
// C# implementation of the approachusing 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 boughtfunction 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> |
5
Time Complexity: O(1) as constant operations done.
Space Complexity: O(1) as no extra space has been used.
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