GCD of more than two (or array) numbers

1.9

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 following.

result = arr[0]
For i = 1 to n-1
   result = GCD(result, arr[i])

Below is C++ implementation of above idea.

C++

// C++ program to find GCD of two or
// more numbers
#include<bits/stdc++.h>
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<n; i++)
        result = gcd(arr[i], result);

    return result;
}

// Driven code
int main()
{
    int arr[] = {2, 4, 6, 8, 16};
    int n = sizeof(arr)/sizeof(arr[0]);
    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 = arr[0];
        for (int i=1; i<n; i++)
            result = gcd(arr[i], result);
        
        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
# This 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
   
    
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


Output:

2

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.

GATE CS Corner    Company Wise Coding Practice

Recommended Posts:



1.9 Average Difficulty : 1.9/5.0
Based on 23 vote(s)










Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.