Skip to content
Related Articles
GCD of more than two (or array) numbers
• Difficulty Level : Easy
• Last Updated : 09 Mar, 2021

Given an array of numbers, find GCD of the array elements. In a previous post we find GCD of two number.
Examples:

```Input  : arr[] = {1, 2, 3}
Output : 1

Input  : arr[] = {2, 4, 6, 8}
Output : 2```

The GCD of three or more numbers equals the product of the prime factors common to all the numbers, but it can also be calculated by repeatedly taking the GCDs of pairs of numbers.

```gcd(a, b, c) = gcd(a, gcd(b, c))
= gcd(gcd(a, b), c)
= gcd(gcd(a, c), b)```

For an array of elements, we do the following. We will also check for the result if the result at any step becomes 1 we will just return the 1 as gcd(1,x)=1.

```result = arr
For i = 1 to n-1
result = GCD(result, arr[i])```

Below is the implementation of the above idea.

## C++

 `// C++ program to find GCD of two or``// more numbers``#include ``using` `namespace` `std;` `// Function to return gcd of a and b``int` `gcd(``int` `a, ``int` `b)``{``    ``if` `(a == 0)``        ``return` `b;``    ``return` `gcd(b % a, a);``}` `// Function to find gcd of array of``// numbers``int` `findGCD(``int` `arr[], ``int` `n)``{``    ``int` `result = arr;``    ``for` `(``int` `i = 1; i < n; i++)``    ``{``        ``result = gcd(arr[i], result);` `        ``if``(result == 1)``        ``{``           ``return` `1;``        ``}``    ``}``    ``return` `result;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 2, 4, 6, 8, 16 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``cout << findGCD(arr, n) << endl;``    ``return` `0;``}`

## JAVA

 `// Java program to find GCD of two or``// more numbers` `public` `class` `GCD {``    ``// Function to return gcd of a and b``    ``static` `int` `gcd(``int` `a, ``int` `b)``    ``{``        ``if` `(a == ``0``)``            ``return` `b;``        ``return` `gcd(b % a, a);``    ``}` `    ``// Function to find gcd of array of``    ``// numbers``    ``static` `int` `findGCD(``int` `arr[], ``int` `n)``    ``{``        ``int` `result = ``0``;``        ``for` `(``int` `element: arr){``            ``result = gcd(result, element);` `            ``if``(result == ``1``)``            ``{``               ``return` `1``;``            ``}``        ``}` `        ``return` `result;``    ``}` `    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``2``, ``4``, ``6``, ``8``, ``16` `};``        ``int` `n = arr.length;``        ``System.out.println(findGCD(arr, n));``    ``}``}` `// This code is contributed by Saket Kumar`

## Python

 `# GCD of more than two (or array) numbers` `# Function implements the Euclidian``# algorithm to find H.C.F. of two number``def` `find_gcd(x, y):``    ` `    ``while``(y):``        ``x, y ``=` `y, x ``%` `y``    ` `    ``return` `x``        ` `# Driver Code       ``l ``=` `[``2``, ``4``, ``6``, ``8``, ``16``]` `num1 ``=` `l[``0``]``num2 ``=` `l[``1``]``gcd ``=` `find_gcd(num1, num2)` `for` `i ``in` `range``(``2``, ``len``(l)):``    ``gcd ``=` `find_gcd(gcd, l[i])``    ` `print``(gcd)` `# Code contributed by Mohit Gupta_OMG`

## C#

 `// C# program to find GCD of``// two or more numbers``using` `System;` `public` `class` `GCD {``    ` `    ``// Function to return gcd of a and b``    ``static` `int` `gcd(``int` `a, ``int` `b)``    ``{``        ``if` `(a == 0)``            ``return` `b;``        ``return` `gcd(b % a, a);``    ``}` `    ``// Function to find gcd of``    ``// array of numbers``    ``static` `int` `findGCD(``int``[] arr, ``int` `n)``    ``{``        ``int` `result = arr;``        ``for` `(``int` `i = 1; i < n; i++){``            ``result = gcd(arr[i], result);` `            ``if``(result == 1)``            ``{``               ``return` `1;``            ``}``        ``}` `        ``return` `result;``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``int``[] arr = { 2, 4, 6, 8, 16 };``        ``int` `n = arr.Length;``        ``Console.Write(findGCD(arr, n));``    ``}``}` `// This code is contributed by nitin mittal`

## PHP

 ``

## Javascript

 ``

Output:

`2`

Time Complexity: O(N * log(M)), where M is the smallest element of the array
Auxiliary Space: O(1)
This article is contributed by DANISH_RAZA . If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

My Personal Notes arrow_drop_up