# Maximum money that can be withdrawn in two steps

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 coins 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `using` `namespace` `std; ` ` `  `// Function to return the maximum coins we can get ` `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 += max(X, Y); ` ` `  `    ``return` `coins; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `X = 7, Y = 5; ` ` `  `    ``cout << maxCoins(X, Y); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `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 approach ` `using` `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

 ` 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. ` ` `  `?> `

Output:

```13
```

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