Suma mínima de los elementos de una array después de restar los elementos más pequeños de los más grandes

Dado un arreglo arr , la tarea es encontrar la suma mínima de los elementos del arreglo después de aplicar la siguiente operación: 
Para cualquier par del arreglo, si a[i] > a[j] entonces a[i] = a[ yo] – a[j] .

Ejemplos: 

Entrada: arr[] = {1, 2, 3} 
Salida: 3 
array modificada será {1, 1, 1}

Entrada: a = {2, 4, 6} 
Salida: 6 
array modificada será {2, 2, 2} 
 

Planteamiento: Observe aquí que después de cada operación, el MCD de todos los elementos seguirá siendo el mismo. Entonces, al final, cada elemento será igual al mcd de todos los elementos de la array después de aplicar la operación dada. 
Entonces, la respuesta final será (n * mcd) . 

A continuación se muestra la implementación del enfoque anterior:  

// 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 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 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# 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 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
?>

<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>

5

Complejidad de tiempo: O(n * log(min(a, b))), donde a y b son dos parámetros del gcd.

Espacio auxiliar: O(log(min(a, b)))