# Minimizing array sum by subtracting larger elements from smaller ones

• Difficulty Level : Medium
• Last Updated : 25 Jul, 2022

Given an array of n elements perform, we need to minimize the array sum. We are allowed to perform the below operation any number of times.

• Choose any two elements from the array say A and B where A > B and then subtract B from A.

Examples:

```
Input : 1
arr[] = 1
Output : 1
There is no need to apply the above operation
because there is only a single number that is 1.
Hence, the minimum sum of the array is 1.

Input : 3
arr[] = 2 4 6
Output : 6
Perform the following operations:-
subtract 2 from 4 then the array becomes 2 2 6
subtract 2 (at 2nd position) from 6 the array
becomes 2 2 4
subtract 2 (at 2nd position) from 4 the array
becomes 2 2 2
Now the sum of the array will be 6.```

Approach:

After applying all the operations, all the values in the given array will be equal otherwise we can still choose two numbers A and B such that A > B and can reduce the sum further. The value of each element after applying all the operations will be equal to the gcd of the array i.e. ans.
Therefore, the minimum possible sum will be equal to n * ans.

Below is the implementation of the above approach:

## C++

 `// CPP program to find the minimum``// sum of the array.``#include ``using` `namespace` `std;` `// returns gcd of two numbers``int` `gcd(``int` `a, ``int` `b)``{``    ``if` `(b == 0)``        ``return` `a;   ``    ``return` `gcd(b, a % b);``}` `// returns the gcd of the array.``int` `gcdofArray(``int` `arr[], ``int` `n)``{``    ``int` `ans = arr;``    ``for` `(``int` `i = 1; i < n; i++)``        ``ans = gcd(ans, arr[i]);   ``    ``return` `ans;``}` `// Driver Function``int` `main()``{``    ``int` `arr[] = { 2, 4, 6 }, n;``    ``n = ``sizeof``(arr) / ``sizeof``(arr);``    ``cout << n * gcdofArray(arr, n)``         ``<< endl;``    ``return` `0;``}`

## Java

 `// java program to find the minimum``// sum of the array.``import` `java.io.*;` `class` `GFG {``    ` `    ``// returns gcd of two numbers``    ``static` `int` `gcd(``int` `a, ``int` `b)``    ``{``        ``if` `(b == ``0``)``            ``return` `a;``        ``return` `gcd(b, a % b);``    ``}``    ` `    ``// returns the gcd of the array.``    ``static` `int` `gcdofArray(``int` `arr[], ``int` `n)``    ``{``        ``int` `ans = arr[``0``];``        ``for` `(``int` `i = ``1``; i < n; i++)``            ``ans = gcd(ans, arr[i]);``        ``return` `ans;``    ``}``    ` `    ``// Driver Function``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `arr[] = { ``2``, ``4``, ``6` `}, n;``        ``n = arr.length;``        ``System.out.println( n * gcdofArray(arr, n)) ;``        ` `    ``}``}` `// This code is contributed by vt_m`

## Python3

 `# Python3 code to find the minimum``# sum of the array.` `# returns gcd of two numbers``def` `gcd(a, b):``    ``if` `b ``=``=` `0``:``        ``return` `a``    ``return` `gcd(b, a ``%` `b)``    ` `# returns the gcd of the array.``def` `gcdofArray(arr, n):``    ``ans ``=` `arr[``0``]``    ``for` `i ``in` `range``(n):``        ``ans ``=` `gcd(ans, arr[i])``    ``return` `ans``    ` `# Driver Code``arr ``=` `[ ``2``, ``4``, ``6` `]``n ``=` `len``(arr)``print``(n ``*` `gcdofArray(arr, n))` `# This code is contributed by "Sharad_Bhardwaj".`

## C#

 `// C# program to find the minimum``// sum of the array.``using` `System;` `class` `GFG``{``    ` `    ``// returns gcd of two numbers``    ``static` `int` `gcd(``int` `a, ``int` `b)``    ``{``        ``if` `(b == 0)``            ``return` `a;``        ``return` `gcd(b, a % b);``    ``}``    ` `    ``// returns the gcd of the array.``    ``static` `int` `gcdofArray(``int` `[]arr, ``int` `n)``    ``{``        ``int` `ans = arr;``        ``for` `(``int` `i = 1; i < n; i++)``            ``ans = gcd(ans, arr[i]);``        ``return` `ans;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main ()``    ``{``        ``int` `[]arr = {2, 4, 6};``        ``int` `n;``        ``n = arr.Length;``        ``Console.WriteLine(n * gcdofArray(arr, n)) ;``        ` `    ``}``}` `// This code is contributed by vt_m.`

## PHP

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## Javascript

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Output

```6
```

Time Complexity: O(N*logN), as we are using a loop to traverse N times and in each traversal, we are calling the gcd function which costs logN time. Where N is the number of elements in the array.
Auxiliary Space: O(logN), due to recursive stack space in gcd function.

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