Minimize the sum of the array according the given condition

Given an array of integers A. The task is to minimise the sum of the elements of the array using the following rule:
Choose two indices i and j and an arbitrary integer x, such that x is a divisor of A[i] and change them as following A[i] = A[i]/x and A[j] = A[j]*x.

Examples:

Input: A = { 1, 2, 3, 4, 5 }
Output: 14



Divide A[3] by 2 then
A[3] = 4/2 = 2,

Multiply A[0] by 2 then
A[0] = 1*2 = 2

Updated array A = { 2, 2, 3, 2, 5 }
Hence sum = 14

Input: A = { 2, 4, 2, 3, 7 }
Output: 18

Approach: If any number is divided by x then it is optimal to multiply the x with the smallest number present in the array.

The idea is to get the minimum of the array and find the divisors of the particular element and keep checking that by how much the sum is reduced.

Below is the implementation of the above approach:

C++

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// C++ implementation
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the minimum sum
void findMin(int arr[], int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
  
    // sort the array to find the
    // minimum element
    sort(arr, arr + n);
  
    int min = arr[0];
    int max = 0;
  
    for (int i = n - 1; i >= 1; i--) {
        int num = arr[i];
        int total = num + min;
        int j;
  
        // finding the number to
        // divide
        for (j = 2; j <= num; j++) {
            if (num % j == 0) {
                int d = j;
                int now = (num / d)
                          + (min * d);
  
                // Checking to what
                // instance the sum
                // has decreased
                int reduce = total - now;
  
                // getting the max
                // difference
                if (reduce > max)
                    max = reduce;
            }
        }
    }
    cout << (sum - max);
}
  
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    findMin(arr, n);
}

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Java

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// Java implementation of the above approach
import java.util.*;
  
class GFG 
{
      
    // Function to return the minimum sum 
    static void findMin(int arr[], int n) 
    
        int sum = 0
        for (int i = 0; i < n; i++) 
            sum += arr[i]; 
      
        // sort the array to find the 
        // minimum element 
        Arrays.sort(arr); 
      
        int min = arr[0]; 
        int max = 0
      
        for (int i = n - 1; i >= 1; i--) 
        
            int num = arr[i]; 
            int total = num + min; 
            int j; 
      
            // finding the number to 
            // divide 
            for (j = 2; j <= num; j++) 
            
                if (num % j == 0
                
                    int d = j; 
                    int now = (num / d) + 
                              (min * d); 
      
                    // Checking to what 
                    // instance the sum 
                    // has decreased 
                    int reduce = total - now; 
      
                    // getting the max 
                    // difference 
                    if (reduce > max) 
                        max = reduce; 
                
            
        
        System.out.println(sum - max); 
    
      
    // Driver Code 
    public static void main (String[] args) 
    
        int arr[] = { 1, 2, 3, 4, 5 }; 
        int n = arr.length; 
        findMin(arr, n); 
    }
}
  
// This code is contributed by AnkitRai01

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Python3

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# Function to return the minimum sum 
def findMin(arr, n):
    sum = 0
    for i in range(0, n): 
        sum = sum + arr[i]
  
    # sort the array to find the 
    # minimum element 
    arr.sort()
  
    min = arr[0]
    max = 0
  
    for i in range(n - 1, 0, -1): 
        num = arr[i]
        total = num + min
  
        # finding the number to 
        # divide 
        for j in range(2, num + 1): 
            if(num % j == 0):
                d = j
                now = (num // d) + (min * d)
  
                # Checking to what 
                # instance the sum 
                # has decreased 
                reduce = total - now
  
                # getting the max 
                # difference 
                if(reduce > max):
                    max = reduce
  
    print(sum - max)
  
# Driver Code 
arr = [1, 2, 3, 4, 5 ]
n = len(arr)
findMin(arr, n)
  
# This code is contributed by Sanjit_Prasad

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C#

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// C# implementation of the above approach
using System;
      
class GFG 
{
      
    // Function to return the minimum sum 
    static void findMin(int []arr, int n) 
    
        int sum = 0; 
        for (int i = 0; i < n; i++) 
            sum += arr[i]; 
      
        // sort the array to find the 
        // minimum element 
        Array.Sort(arr); 
      
        int min = arr[0]; 
        int max = 0; 
      
        for (int i = n - 1; i >= 1; i--) 
        
            int num = arr[i]; 
            int total = num + min; 
            int j; 
      
            // finding the number to 
            // divide 
            for (j = 2; j <= num; j++) 
            
                if (num % j == 0) 
                
                    int d = j; 
                    int now = (num / d) + 
                              (min * d); 
      
                    // Checking to what 
                    // instance the sum 
                    // has decreased 
                    int reduce = total - now; 
      
                    // getting the max 
                    // difference 
                    if (reduce > max) 
                        max = reduce; 
                
            
        
        Console.WriteLine(sum - max); 
    
      
    // Driver Code 
    public static void Main (String[] args) 
    
        int []arr = { 1, 2, 3, 4, 5 }; 
        int n = arr.Length; 
        findMin(arr, n); 
    }
}
  
// This code is contributed by PrinciRaj1992

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Output:

14


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