There are two cash lockers, one has X number of coins and the other has Y number of coins, you can withdraw money at max two times, when you withdraw from a locker you will get the total money of the locker and the locker will be refilled with Z – 1 coin if it had Z coins initially. The task is to find the maximum coins you can get.
Examples:
Input: X = 6, Y = 3
Output: 11
Take from locker X i.e. 6
Now, X = 5 and Y = 3
Take again from locker X i.e. 5.
Input: X = 4, Y = 4
Output: 8
Approach: In order to maximize the number of coins, take from the locker which has the maximum value then update the locker and again take from the locker with the maximum value.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the maximum coins we can getint maxCoins(int X, int Y){ // Update elements such that X > Y if (X < Y) swap(X, Y); // Take from the maximum int coins = X; // Refill X--; // Again, take the maximum coins += max(X, Y); return coins;}// Driver codeint main(){ int X = 7, Y = 5; cout << maxCoins(X, Y); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG { // Function to return the maximum coins we can get static int maxCoins(int X, int Y) { // Update elements such that X > Y if (X < Y) { swap(X, Y); } // Take from the maximum int coins = X; // Refill X--; // Again, take the maximum coins += Math.max(X, Y); return coins; } static void swap(int X, int Y) { int temp = X; X = Y; Y = temp; } // Driver code public static void main(String[] args) { int X = 7, Y = 5; System.out.println(maxCoins(X, Y)); }}// This code has been contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach # Function to return the maximum coins we can get def maxCoins(X, Y) : # Update elements such that X > Y if (X < Y) : X, Y = Y, X; # Take from the maximum coins = X; # Refill X -= 1; # Again, take the maximum coins += max(X, Y); return coins; # Driver code if __name__ == "__main__" : X = 7; Y = 5; print(maxCoins(X, Y)); # This code is contributed by Ryuga |
C#
// C# implementation of the approachusing System;class GFG { // Function to return the maximum coins we can get static int maxCoins(int X, int Y) { // Update elements such that X > Y if (X < Y) { swap(X, Y); } // Take from the maximum int coins = X; // Refill X--; // Again, take the maximum coins += Math.Max(X, Y); return coins; } static void swap(int X, int Y) { int temp = X; X = Y; Y = temp; } // Driver code public static void Main(String[] args) { int X = 7, Y = 5; Console.WriteLine(maxCoins(X, Y)); }}/* This code contributed by PrinciRaj1992 */ |
PHP
<?php// PHP implementation of the approach // Function to return the maximum coins we can get function maxCoins($X, $Y) { // Update elements such that X > Y if ($X < $Y) swap($X, $Y); // Take from the maximum $coins = $X; // Refill $X--; // Again, take the maximum $coins += max($X, $Y); return $coins; } // Driver code $X = 7;$Y = 5; echo maxCoins($X, $Y); // This code is contributed by Naman_Garg.?> |
Javascript
<script>// Javascript implementation of the approach// Function to return the maximum coins we can getfunction maxCoins(X, Y){ // Update elements such that X > Y if (X < Y) { let temp = X; X = Y; Y = temp; } // Take from the maximum let coins = X; // Refill X--; // Again, take the maximum coins += Math.max(X, Y); return coins;}// Driver code let X = 7, Y = 5; document.write(maxCoins(X, Y));</script> |
13
Time Complexity: O(1), as there is no loop.
Auxiliary Space: O(1), as no extra space has been taken.
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