Related Articles

# Minimizing array sum by subtracting larger elements from smaller ones

• Difficulty Level : Medium
• Last Updated : 01 Apr, 2021

Given an array of n elements perform, we need to minimize array sum. We are allowed to perform 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. Value of each element after applying all the operation will be equal to gcd of the array i.e. ans.
Therefore, minimum possible sum will be equal to n * ans.
Below is the implementation of 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

 ``

## Javascript

 ``

Output:

`6`

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up