# 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}

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

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 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



Output:

5

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