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Minimizing array sum by subtracting larger elements from smaller ones

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  • Difficulty Level : Medium
  • Last Updated : 25 Jul, 2022
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Given an array of n elements perform, we need to minimize the array sum. We are allowed to perform the 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. The value of each element after applying all the operations will be equal to the gcd of the array i.e. ans. 
Therefore, the minimum possible sum will be equal to n * ans.

Below is the implementation of the above approach: 

C++




// CPP program to find the minimum
// sum of the array.
#include <bits/stdc++.h>
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[0];
    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[0]);
    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[0];
        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




<?php
// PHP program to find the minimum
// sum of the array.
 
// returns gcd of two numbers
function gcd($a, $b)
{
    if ($b == 0)
        return $a;
    return gcd($b, $a % $b);
}
 
// returns the gcd of the array.
function gcdofArray($arr, $n)
{
    $ans = $arr[0];
    for ($i = 1; $i < $n; $i++)
        $ans = gcd($ans, $arr[$i]);
    return $ans;
}
 
// Driver Code
$arr = array( 2, 4, 6 );
$n = count($arr);
echo $n * gcdofArray($arr, $n), "\n";
 
// This code is contributed by Sach
?>

Javascript




<script>
 
    // Javascript program to find the minimum
    // sum of the array.
     
    // returns gcd of two numbers
    function gcd(a, b)
    {
        if (b == 0)
            return a;   
        return gcd(b, a % b);
    }
 
    // returns the gcd of the array.
    function gcdofArray(arr, n)
    {
        let ans = arr[0];
        for (let i = 1; i < n; i++)
            ans = gcd(ans, arr[i]);   
        return ans;
    }
 
    let arr = [ 2, 4, 6 ], n;
    n = arr.length;
    document.write(n * gcdofArray(arr, n));
 
</script>

Output

6

 Time Complexity: O(N*logN), as we are using a loop to traverse N times and in each traversal, we are calling the gcd function which costs logN time. Where N is the number of elements in the array.
Auxiliary Space: O(logN), due to recursive stack space in gcd function.


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