Minimum sum of the elements of an array after subtracting smaller elements from larger

Given an array arr, the task is to find the minimum sum of the elements of the array after applying the following operation: 
For any pair from the array, if a[i] > a[j] then a[i] = a[i] – a[j].

Examples:

Input: arr[] = {1, 2, 3} 
Output: 3 
modified array will be {1, 1, 1}

Input: a = {2, 4, 6} 
Output: 6 
modified array will be {2, 2, 2} 

Approach: Observe here that after each operation, the GCD of all the elements will remain the same. So, in the end, every element will be equal to the gcd of all the elements of the array after applying the given operation. 
So, the final answer will be (n * gcd). 

Below is the implementation of the above approach:  

C++

// CPP program to Find the minimum sum
// of given array after applying given operation.
#include <bits/stdc++.h>
using namespace std;

// Function to Find the minimum sum
// of given array after applying given operation.
int MinSum(int a[], int n)
{
    // to store final gcd value
    int gcd = a[0];

    // get gcd of the whole array
    for (int i = 1; i < n; i++)
        gcd = __gcd(a[i], gcd);

    return n * gcd;
}

// Driver code
int main()
{

    int a[] = { 20, 14, 6, 8, 15 };

    int n = sizeof(a) / sizeof(a[0]);

    // function call
    cout << MinSum(a, n);

    return 0;
}

Java

// Java program to Find the minimum sum
// of given array after applying given operation.

import java.io.*;

class GFG {

// Recursive function to return gcd of a and b
static int __gcd(int a, int b)
{
    // Everything divides 0
    if (a == 0)
       return b;
    if (b == 0)
       return a;

    // base case
    if (a == b)
        return a;

    // a is greater
    if (a > b)
        return __gcd(a-b, b);
    return __gcd(a, b-a);
}
// Function to Find the minimum sum
// of given array after applying given operation.
static int MinSum(int []a, int n)
{
    // to store final gcd value
    int gcd = a[0];

    // get gcd of the whole array
    for (int i = 1; i < n; i++)
        gcd = __gcd(a[i], gcd);

    return n * gcd;
}

// Driver code

    public static void main (String[] args) {
            int a[] = { 20, 14, 6, 8, 15 };

    int n = a.length;

    // function call
    System.out.println(MinSum(a, n));
    }
}
// This code is contributed by anuj_67..

Python3

# Python3 program to Find the minimum
# sum of given array after applying
# given operation.
import math

# Function to Find the minimum sum
# of given array after applying
# given operation.
def MinSum(a, n):

    # to store final gcd value
    gcd = a[0]

    # get gcd of the whole array
    for i in range(1, n):
        gcd = math.gcd(a[i], gcd)

    return n * gcd

# Driver code
if __name__ == "__main__":

    a = [20, 14, 6, 8, 15 ]

    n = len(a)

    # function call
    print(MinSum(a, n))

# This code is contributed by ita_c

C#

// C# program to Find the minimum sum
// of given array after applying given operation.

using System;
class GFG {

    // Recursive function to return gcd of a and b
    static int __gcd(int a, int b)
    {
        // Everything divides 0
        if (a == 0)
           return b;
        if (b == 0)
           return a;

        // base case
        if (a == b)
            return a;

        // a is greater
        if (a > b)
            return __gcd(a-b, b);
        return __gcd(a, b-a);
    }

    // Function to Find the minimum sum
    // of given array after applying given operation.
    static int MinSum(int []a, int n)
    {
        // to store final gcd value
        int gcd = a[0];

        // get gcd of the whole array
        for (int i = 1; i < n; i++)
            gcd = __gcd(a[i], gcd);

        return n * gcd;
    }


    // Driver Program to test above function
    static void Main()
    {
        int []a = { 20, 14, 6, 8, 15 };
        int n = a.Length;
        Console.WriteLine(MinSum(a, n));
    }

    // This code is contributed by Ryuga.
}

PHP

<?php
// PHP program to Find the minimum sum of
// given array after applying given operation.

// Function to Find the minimum sum
// of given array after applying
// given operation.
function gcd($a, $b)
{
    if ($b == 0)
        return $a;
    return gcd($b, $a % $b);

}

function MinSum($a, $n)
{
    // to store final gcd value
    $gcdd = $a[0];

    // get gcd of the whole array
    for ($i = 1; $i < $n; $i++)
        $gcdd = gcd($a[$i], $gcdd);

    return $n * $gcdd;
}

// Driver code
$a = array( 20, 14, 6, 8, 15 );

$n = count($a);

// function call
echo MinSum($a, $n);

// This code is contributed by mits
?>

Javascript

<script>

// Javascript program to Find the minimum sum
// of given array after applying given operation.

// Recursive function to return gcd of a and b
function __gcd(a, b)
{

    // Everything divides 0
    if (a == 0)
       return b;
    if (b == 0)
       return a;

    // base case
    if (a == b)
        return a;

    // a is greater
    if (a > b)
        return __gcd(a - b, b);
    return __gcd(a, b - a);
}

// Function to Find the minimum sum
// of given array after applying
// given operation.
function MinSum(a, n)
{

    // To store final gcd value
    var gcd = a[0];

    // Get gcd of the whole array
    for(var i = 1; i < n; i++)
        gcd = __gcd(a[i], gcd);

    return n * gcd;
}

// Driver code
var a = [ 20, 14, 6, 8, 15 ];
var n = a.length;

// Function call
document.write( MinSum(a, n));

// This code is contributed by noob2000

</script>

Output:

5

Time Complexity: O(n * log(min(a, b))), where a and b are two parameters of the gcd.

Auxiliary Space: O(log(min(a, b)))