# GCD of more than two (or array) numbers

• Difficulty Level : Easy
• Last Updated : 16 Nov, 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```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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[0]
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 &lt;bits/stdc++.h&gt;
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[0];
for (int i = 1; i &lt; 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[0]);
cout &lt;&lt; findGCD(arr, n) &lt;&lt; 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[0];
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

```<?php
// PHP program to find GCD of two or
// more numbers

// Function to return gcd of a and b
function gcd( \$a, \$b)
{
if (\$a == 0)
return \$b;
return gcd(\$b % \$a, \$a);
}

// Function to find gcd of array of
// numbers
function findGCD(\$arr, \$n)
{
\$result = \$arr[0];

for (\$i = 1; \$i &lt; \$n; \$i++){
\$result = gcd(\$arr[\$i], \$result);

if(\$result == 1)
{
return 1;
}
}
return \$result;
}

// Driver code
\$arr = array( 2, 4, 6, 8, 16 );
\$n = sizeof(\$arr);
echo (findGCD(\$arr, \$n));

// This code is contributed by
?>
```

## Javascript

```<script>

// Javascript program to find GCD of two or
// more numbers

// Function to return gcd of a and b
function gcd(a, b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}

// Function to find gcd of array of
// numbers
function findGCD(arr, n)
{
let result = arr[0];
for (let i = 1; i < n; i++)
{
result = gcd(arr[i], result);

if(result == 1)
{
return 1;
}
}
return result;
}

// Driver code

let arr = [ 2, 4, 6, 8, 16 ];
let n = arr.length;
document.write(findGCD(arr, n) + "<br>");

// This is code is contributed by Mayank Tyagi

</script>
```

Output:

`2`

Time Complexity: O(N * log(N)), where N is the largest element of the array
Auxiliary Space: O(1)

Recursive Method: Implementation of Algorithm recursively :

## C++

```#include <bits/stdc++.h>
using namespace std;

// recursive implementation
int GcdOfArray(vector<int>& arr, int idx)
{
if (idx == arr.size() - 1) {
return arr[idx];
}
int a = arr[idx];
int b = GcdOfArray(arr, idx + 1);
return __gcd(
a, b); // __gcd(a,b) is inbuilt library function
}

int main()
{
vector<int> arr = { 1, 2, 3 };
cout << GcdOfArray(arr, 0) << "\n";

arr = { 2, 4, 6, 8 };
cout << GcdOfArray(arr, 0) << "\n";
return 0;
}```

## Java

```import java.util.*;
class GFG
{

// Recursive function to return gcd of a and b
static int __gcd(int a, int b)
{
return b == 0? a:__gcd(b, a % b);
}

// recursive implementation
static int GcdOfArray(int[] arr, int idx)
{
if (idx == arr.length - 1) {
return arr[idx];
}
int a = arr[idx];
int b = GcdOfArray(arr, idx + 1);
return __gcd(
a, b); // __gcd(a,b) is inbuilt library function
}

public static void main(String[] args)
{
int[] arr = { 1, 2, 3 };
System.out.print(GcdOfArray(arr, 0)+ "\n");

int[] arr1 = { 2, 4, 6, 8 };
System.out.print(GcdOfArray(arr1, 0)+ "\n");
}
}

// This code is contributed by gauravrajput1
```
Output
```1
2```

Time Complexity: O(N * log(N)), where N is the largest element of the array

Auxiliary Space: O(N)

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