Find the final number obtained after performing the given operation

Given an array of positive distinct integers arr[], the task is to find the final number obtained by performing the following operation on the elements of the array: 
Operation: Take two unequal numbers and replace the larger number with their difference until all numbers become equal.
Examples:

Input: arr[] = {5, 2, 3} 
Output: 1 
5 – 3 = 2, arr[] = {2, 2, 3} 
3 – 2 = 1, arr[] = {2, 2, 1} 
2 – 1 = 1, arr[] = {2, 1, 1} 
2 – 1 = 1, arr[] = {1, 1, 1}
Input: arr[] = {3, 9, 6, 36} 
Output: 3 

Naive approach: Since final answer will always be distinct, one can just sort the array and replace the largest term with the difference of the two largest elements and repeat the process until all the numbers become equal.

C++

#include <algorithm>
#include <iostream>

int finalNumber(int arr[], int n)
{
    std::sort(arr,
              arr + n); // Sort the array in ascending order

    while (arr[n - 1] != arr[0]) { // Keep looping until all
                                   // elements are equal
        int diff = arr[n - 1] - arr[0];
        arr[n - 1] = diff;
        std::sort(arr,
                  arr + n); // Sort the array again after
                            // changing the last element
    }

    return arr[0];
}

int main()
{
    int arr[] = { 3, 9, 6, 36 };
    int n = sizeof(arr)
            / sizeof(arr[0]); // Get the size of the array

    std::cout << finalNumber(arr, n) << std::endl;
    return 0;
}

Java

import java.util.Arrays;

public class Main {
  public static int finalNumber(int[] arr) {
    int n = arr.length;
    Arrays.sort(arr);
    while (arr[n - 1] != arr[0]) {
      int diff = arr[n - 1] - arr[0];
      arr[n - 1] = diff;
      Arrays.sort(arr);
    }
    return arr[0];
  }

  public static void main(String[] args) {

    int[] arr2 = {3, 9, 6, 36};

    System.out.println(finalNumber(arr2));
  }
}

Python3

def final_number(arr):
    n = len(arr)  # Get the length of the input array
    arr.sort()  # Sort the array using sort function
    while arr[n - 1] != arr[0]:  # While the largest and smallest elements in the array are not equal
        diff = arr[n - 1] - arr[0]  # Calculate the difference between the largest and smallest elements
        arr[n - 1] = diff  # Replace the largest element with the difference
        arr.sort()  # Sort the array again
    return arr[0]  # Return the smallest element in the array

if __name__ == '__main__':
    arr = [3, 9, 6, 36]  # Define the input array
    print(final_number(arr))  # Call the final_number function and print the result

Javascript

function finalNumber(arr) {
  const n = arr.length; // Get the length of the input array
  arr.sort((a, b) => a - b); // Sort the array using sort function

  while (arr[n - 1] !== arr[0]) { // While the largest and smallest elements in the array are not equal
    const diff = arr[n - 1] - arr[0]; // Calculate the difference between the largest and smallest elements
    arr[n - 1] = diff; // Replace the largest element with the difference
    arr.sort((a, b) => a - b); // Sort the array again
  }

  return arr[0]; // Return the smallest element in the array
}

const arr = [3, 9, 6, 36]; // Define the input array
console.log(finalNumber(arr)); // Call the finalNumber function and print the result

C#

// C# implementation of the approach
using System;
using System.Linq;

class Program
{
    static int FinalNumber(int[] arr, int n)
    {
        Array.Sort(arr); // Sort the array in ascending order

        while (arr[n - 1] != arr[0]) // Keep looping until all elements are equal
        {
            int diff = arr[n - 1] - arr[0];
            arr[n - 1] = diff;
            Array.Sort(arr); // Sort the array again after changing the last element
        }

        return arr[0];
    }

    static void Main(string[] args)
    {
        int[] arr = { 3, 9, 6, 36 };
        int n = arr.Length;

        Console.WriteLine(FinalNumber(arr, n));
    }
}

// Contributed by adityasharmadev01

Output : 

3 

Efficient approach: From Euclidean’s algorithm, it is known that gcd(a, b) = gcd(a – b, b). This can be extended to gcd(A₁, A₂, A₃, …, A_n) = gcd(A₁ – A₂, A₂, A₃, …, A_n). 
Also, let’s say that after applying the given operation, the final number obtained be K. Hence, from the extended algorithm, it can be said that gcd(A₁, A₂, A₃, …, A_n) = gcd(K, K, …, n times). Since gcd(K, K, …, n times) = K, the solution of the given problem can be found 
by finding the gcd of all the elements of the array.
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the final number
// obtained after performing the
// given operation
int finalNum(int arr[], int n)
{

    // Find the gcd of the array elements
    int result = 0;
    for (int i = 0; i < n; i++) {
        result = __gcd(result, arr[i]);
    }
    return result;
}

// Driver code
int main()
{
    int arr[] = { 3, 9, 6, 36 };
    int n = sizeof(arr) / sizeof(arr[0]);

    cout << finalNum(arr, n);

    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
class GFG
{

// Function to return the final number
// obtained after performing the
// given operation
static int finalNum(int arr[], int n)
{

    // Find the gcd of the array elements
    int result = 0;
    for (int i = 0; i < n; i++)
    {
        result = __gcd(result, arr[i]);
    }
    return result;
}

static int __gcd(int a, int b)
{
    return b == 0? a:__gcd(b, a % b);
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { 3, 9, 6, 36 };
    int n = arr.length;

    System.out.print(finalNum(arr, n));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
from math import gcd as __gcd

# Function to return the final number
# obtained after performing the
# given operation
def finalNum(arr, n):

    # Find the gcd of the array elements
    result = arr[0]
    for i in arr:
        result = __gcd(result, i)
    return result

# Driver code
arr = [3, 9, 6, 36]
n = len(arr)

print(finalNum(arr, n))

# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to return the readonly number
// obtained after performing the
// given operation
static int finalNum(int []arr, int n)
{

    // Find the gcd of the array elements
    int result = 0;
    for (int i = 0; i < n; i++)
    {
        result = __gcd(result, arr[i]);
    }
    return result;
}

static int __gcd(int a, int b)
{
    return b == 0 ? a : __gcd(b, a % b);
}

// Driver code
public static void Main(String[] args)
{
    int []arr = { 3, 9, 6, 36 };
    int n = arr.Length;

    Console.Write(finalNum(arr, n));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>
// javascript implementation of the approach
// Function to return the final number
    // obtained after performing the
    // given operation
    function finalNum(arr , n) {

        // Find the gcd of the array elements
        var result = 0;
        for (i = 0; i < n; i++) {
            result = __gcd(result, arr[i]);
        }
        return result;
    }

    function __gcd(a , b) {
        return b == 0 ? a : __gcd(b, a % b);
    }

    // Driver code

        var arr = [ 3, 9, 6, 36 ];
        var n = arr.length;

        document.write(finalNum(arr, n));

// This code contributed by aashish1995
</script>

Output:

3

Time Complexity: O(N*logN), as we are using a loop to traverse N times so it will cost us O(N) time and _gcd function will take logN time.
Auxiliary Space: O(1), as we are not using any extra space.