# Find maximum array sum after making all elements same with repeated subtraction

Given an array of n elements, find out the maximum possible sum of all elements such that all the elements are equal. Only operation allowed is choosing any two elements and replacing the larger of them by the absolute difference of the two.

Examples:

```Input : 9 12 3 6
Output : 12
Explanation :
9 12 3 6
replace a2 = 12 with a2-a4 = 12 - 6 => 6
i.e, 9 6 3 6

replace a4 = 6 with a4-a3 = 6 - 3 => 3
i.e, 9 6 3 3

replace a1 = 9 with a1-a2 = 9 - 6 => 3
i.e, 3 6 3 3

replace a2 = 6 with a2-a4 = 6 - 3 => 3
i,e. 3 3 3 3

Now, at this point we have all the elements equal,
hence we can return our answer from here.

Input : 4 8 6 10
Output : 8
Explanation :
Resultant array formed will be:
4 8 6 10
replace a4 = 10 with a4-a1 = 10 - 4 => 6
i.e, 4 8 6 6

replace a3 = 6 with a3-a1 = 6 - 4 => 2
i.e, 4 8 2 6

replace a2 = 8 with a2-a4 = 8 - 6 => 2
i.e, 4 2 2 6

replace a4 = 6 with a4-a1 = 6 - 4 => 2
i,e. 4 2 2 2

replace a1 = 4 with a1-a2 = 4 - 2 => 2
i,e. 2 2 2 2

Now, at this point we have all the elements equal,
hence we can return our answer from here.
```

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

By analyzing the given operation i.e,

```    ai = ai - aj          where ai > aj
```

We see that this is similar to finding GCD through Euclidean algorithm as:

`   GCD(a, b) = GCD(b, a - b)`

And also, the order of rearrangement does not matter, we can proceed by taking any two elements and replace the larger value by the absolute difference of the two, and repeat among them till the difference comes out to be zero[both the elements be same]. That is, taking out the GCD of any two numbers. The reason for this to work is, GCD is associative and commutative.

So the idea is the take the GCD of all the elements at once and replace all the elements by that result.

## C++

 `// Maximum possible sum of array after repeated ` `// subtraction operation. ` `#include ` `using` `namespace` `std; ` ` `  `int` `GCD(``int` `a, ``int` `b) ` `{ ` `    ``if` `(b == 0) ` `        ``return` `a; ` `    ``return` `GCD(b, a % b); ` `} ` ` `  `int` `findMaxSumUtil(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `finalGCD = arr; ` `    ``for` `(``int` `i = 1; i < n; i++) ` `        ``finalGCD = GCD(arr[i], finalGCD); ` ` `  `    ``return` `finalGCD; ` `} ` ` `  `// This function basically calls findMaxSumUtil() ` `// to find GCD of all array elements, then it returns ` `// GCD * (Size of array) ` `int` `findMaxSum(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `maxElement = findMaxSumUtil(arr, n); ` `    ``return` `(maxElement * n); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = {8, 20, 12, 36}; ` `    ``int` `n = ``sizeof``(arr)/``sizeof``(arr); ` `    ``cout << findMaxSum(arr, n) << endl; ` `    ``return` `0; ` `} `

## Java

 `// Maximum possible sum of array after repeated ` `// subtraction operation. ` ` `  `import` `java.io.*; ` ` `  `class` `GFG { ` ` `  `    ``static` `int` `GCD(``int` `a, ``int` `b) ` `    ``{ ` `        ``if` `(b == ``0``) ` `            ``return` `a; ` `        ``return` `GCD(b, a % b); ` `    ``} ` `     `  `    ``static` `int` `findMaxSumUtil(``int` `arr[], ``int` `n) ` `    ``{ ` `        ``int` `finalGCD = arr[``0``]; ` `        ``for` `(``int` `i = ``1``; i < n; i++) ` `            ``finalGCD = GCD(arr[i], finalGCD); ` `     `  `        ``return` `finalGCD; ` `    ``} ` `     `  `    ``// This function basically calls  ` `    ``// findMaxSumUtil() to find GCD of all  ` `    ``// array elements, then it returns ` `    ``// GCD * (Size of array) ` `    ``static` `int` `findMaxSum(``int` `arr[], ``int` `n) ` `    ``{ ` `        ``int` `maxElement = findMaxSumUtil(arr, n); ` `        ``return` `(maxElement * n); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args) { ` `         `  `        ``int` `arr[] = {``8``, ``20``, ``12``, ``36``}; ` `        ``int` `n = arr.length; ` `         `  `        ``System.out.println(findMaxSum(arr, n)); ` `    ``} ` `} ` ` `  `//This code is contributed by vt_m. `

## Python3

 `# Maximum possible sum of array after  ` `# repeated subtraction operation. ` ` `  `def` `GCD(a, b): ` ` `  `    ``if` `(b ``=``=` `0``): ``return` `a ` `    ``return` `GCD(b, a ``%` `b) ` ` `  `def` `findMaxSumUtil(arr, n): ` ` `  `    ``finalGCD ``=` `arr[``0``] ` `    ``for` `i ``in` `range``(``1``, n): ` `        ``finalGCD ``=` `GCD(arr[i], finalGCD) ` ` `  `    ``return` `finalGCD ` ` `  `# This function basically calls ` `# findMaxSumUtil() to find GCD of ` `# all array elements, then it returns ` `# GCD * (Size of array) ` `def` `findMaxSum(arr, n): ` ` `  `    ``maxElement ``=` `findMaxSumUtil(arr, n) ` `    ``return` `(maxElement ``*` `n) ` ` `  `# Driver code ` `arr ``=` `[``8``, ``20``, ``12``, ``36``] ` `n ``=` `len``(arr) ` `print``(findMaxSum(arr, n)) ` ` `  `# This code is contributed by Anant Agarwal. `

## C#

 `// C# Code for Maximum possible sum of array  ` `// after repeated subtraction operation. ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``static` `int` `GCD(``int` `a, ``int` `b) ` `    ``{ ` `        ``if` `(b == 0) ` `            ``return` `a; ` `        ``return` `GCD(b, a % b); ` `    ``} ` `     `  `    ``static` `int` `findMaxSumUtil(``int` `[]arr, ``int` `n) ` `    ``{ ` `        ``int` `finalGCD = arr; ` `        ``for` `(``int` `i = 1; i < n; i++) ` `            ``finalGCD = GCD(arr[i], finalGCD); ` `     `  `        ``return` `finalGCD; ` `    ``} ` `     `  `    ``// This function basically calls  ` `    ``// findMaxSumUtil() to find GCD of all  ` `    ``// array elements, then it returns ` `    ``// GCD * (Size of array) ` `    ``static` `int` `findMaxSum(``int` `[]arr, ``int` `n) ` `    ``{ ` `        ``int` `maxElement = findMaxSumUtil(arr, n); ` `        ``return` `(maxElement * n); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main () { ` `         `  `        ``int` `[]arr = {8, 20, 12, 36}; ` `        ``int` `n = arr.Length; ` `         `  `        ``Console.WriteLine(findMaxSum(arr, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` `

Output:

``` 16
```

This article is contributed by Shubham Gupta. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : vt_m

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