# 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:
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:

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 ` `#include `   `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, …, An) = gcd(A1 – A2, A2, A3, …, An)
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, …, An) = 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 ` `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

 ``

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.