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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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..

chevron_right


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#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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.
}

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?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
?>

chevron_right


Output:

5


My Personal Notes arrow_drop_up

Coder Machine Learner Social Activist Vocalist

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.