GCD of more than two (or array) numbers

Last Updated : 6 Apr, 2026

Given an array arr[] of non-negative numbers, find GCD of all the array elements. In a previous post we find GCD of two number.

Examples:

Input: arr[] = [1, 2, 3]
Output: 1
Explanation: The GCD of 1, 2, and 3 is 1 because they have no common divisor greater than 1.

Input: arr[] = [2, 4, 6, 8]
Output: 2
Explanation: The GCD of 2, 4, 6, and 8 is 2 because 2 is the greatest number that divides all elements exactly.

Try It Yourself
redirect icon

Table of Content

Using Recursive GCD

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)

We traverse the array while keeping track of a variable that stores the GCD of all the elements processed up to that point. During the ith iteration, we update this GCD by calculating the GCD of the current element and the GCD obtained so far.

We will also check for the result if the result at any step becomes 1 then return 1 as gcd(1, any number) = 1.

C++ 
#include <iostream>
#include <vector>
using namespace std;

// Recursive 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(vector<int> &arr) {
    int res = arr[0];
    for (int i = 1; i < arr.size(); i++) {
        res = gcd(arr[i], res);

        // If res becomes 1 at any iteration then it remains 1
        // So no need to check further
        if (res == 1)
            return 1;
    }
    return res;
}

int main() {
    vector<int> arr = {2, 4, 6, 8, 16};

    cout << findGCD(arr) << endl;
    return 0;
}
C 
#include <stdio.h>

// Recursive 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 array[], int n) {
    int res = array[0];
    for (int i = 1; i < n; i++) {
        res = gcd(array[i], res);

        // If res becomes 1 at any iteration then it remains 1
        // So no need to check further
        if (res == 1)
            return 1;
    }
    return res;
}

int main() {
    int array[] = {2, 4, 6, 8, 16};
    int n = sizeof(array) / sizeof(array[0]);

    printf("%d\n", findGCD(array, n));
    return 0;
}
Java 
import java.util.*;

public class GFG {

    // Recursive function to return gcd of a and b
    public 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
    public static int findGCD(int[] array) {
        int res = array[0];
        for (int i = 1; i < array.length; i++) {
            res = gcd(array[i], res);

            // If res becomes 1 at any iteration then it remains 1
            // So no need to check further
            if (res == 1)
                return 1;
        }
        return res;
    }

    public static void main(String[] args) {
        int[] array = {2, 4, 6, 8, 16};

        System.out.println(findGCD(array));
    }
}
Python 
from functools import reduce

# Recursive function to return gcd of a and b
def gcd(a, b):
    if a == 0:
        return b
    return gcd(b % a, a)


# Function to find gcd of array of numbers
def findGCD(array):
    res = array[0]
    for num in array[1:]:
        res = gcd(num, res)

        # If res becomes 1 at any iteration then it remains 1
        # So no need to check further
        if res == 1:
            return 1
    return res

if __name__ == "__main__":
    array = [2, 4, 6, 8, 16]

    print(findGCD(array))
C# 
using System;
using System.Collections.Generic;

class GFG {

    // Recursive function to return gcd of a and b
    public 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
    public static int findGCD(int[] array) {
        int res = array[0];
        for (int i = 1; i < array.Length; i++) {
            res = gcd(array[i], res);

            // If res becomes 1 at any iteration then it remains 1
            // So no need to check further
            if (res == 1)
                return 1;
        }
        return res;
    }

    static void Main(string[] args) {
        int[] array = {2, 4, 6, 8, 16};

        Console.WriteLine(findGCD(array));
    }
}
JavaScript 
// Recursive 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(array) {
    let res = array[0];
    for (let i = 1; i < array.length; i++) {
        res = gcd(array[i], res);

        // If res becomes 1 at any iteration then it remains 1
        // So no need to check further
        if (res === 1)
            return 1;
    }
    return res;
}

// Driver Code
const array = [2, 4, 6, 8, 16];
console.log(findGCD(array));

Output
2

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

Iterative Method

C++ 
#include <iostream>
#include <vector>
using namespace std;

// Iterative implementation
int getGCD(int a, int b) {
    while (a > 0 && b > 0) {
        if (a > b) {
            a = a % b;
        }
        else {
            b = b % a;
        }
    }
    if (a == 0) {
        return b;
    }
    return a;
}

int GcdOfArray(vector<int>& arr) {
    int gcd = arr[0];
    for (int i = 1; i < arr.size(); i++) {
        gcd = getGCD(gcd, arr[i]);
    }
    return gcd;
}

int main() {
    vector<int> arr = { 2, 4, 6, 8 };
    cout << GcdOfArray(arr) << "\n";
    return 0;
}
C 
#include <stdio.h>

// Iterative implementation
int getGCD(int a, int b) {
    while (a > 0 && b > 0) {
        if (a > b) {
            a = a % b;
        } else {
            b = b % a;
        }
    }
    return (a == 0) ? b : a;
}

int GcdOfArray(int arr[], int n) {
    int gcd = arr[0];
    for (int i = 1; i < n; i++) {
        gcd = getGCD(gcd, arr[i]);
    }
    return gcd;
}

int main() {
    int arr[] = {2, 4, 6, 8};
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("%d\n", GcdOfArray(arr, n));
    return 0;
}
Java 
import java.util.Arrays;
import java.util.List;

class GFG {
  
    // Iterative implementation
    static int getGCD(int a, int b) {
        while (a > 0 && b > 0) {
            if (a > b) {
                a = a % b;
            } else {
                b = b % a;
            }
        }
        return (a == 0) ? b : a;
    }

    static int GcdOfArray(int[] arr) {
        int gcd = arr[0];
        for (int num : arr) {
            gcd = getGCD(gcd, num);
        }
        return gcd;
    }

    public static void main(String[] args) {
        int[] arr = { 2, 4, 6, 8 };
        System.out.println(GcdOfArray(arr));
    }
}
Python 
def getGcd(a, b):
    while a > 0 and b > 0:
        if a > b:
            a = a % b
        else:
            b = b % a
    return b if a == 0 else a

# Function to find GCD of an array
def GcdOfArray(arr):
    gcd = arr[0]
    for num in arr:
        gcd = getGcd(gcd, num)
    return gcd

if __name__ == '__main__':
    arr = [2, 4, 6, 8]
    print(GcdOfArray(arr))
C# 
using System;
using System.Linq;

class GFG {
  
    // Iterative implementation
    static int getGCD(int a, int b) {
        while (a > 0 && b > 0) {
            if (a > b) {
                a = a % b;
            } else {
                b = b % a;
            }
        }
        return (a == 0) ? b : a;
    }

    static int GcdOfArray(int[] arr) {
        int gcd = arr[0];
        foreach (int num in arr) {
            gcd = getGCD(gcd, num);
        }
        return gcd;
    }

    public static void Main() {
        int[] arr = {2, 4, 6, 8};
        Console.WriteLine(GcdOfArray(arr));
    }
}
JavaScript 
function getGCD(a, b) {
    while (a > 0 && b > 0) {
        if (a > b) {
            a = a % b;
        } else {
            b = b % a;
        }
    }
    return (a === 0) ? b : a;
}

function GcdOfArray(arr) {
    let gcd = 0;
    for (let num of arr) {
        gcd = getGCD(gcd, num);
    }
    return gcd;
}

// Driver Code
const arr = [2, 4, 6, 8];
console.log(GcdOfArray(arr));

Output
2

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

Using Built-in Methods

Languages like C++, Java, Python, C# etc., have built-in methods for calculating the GCD of two numbers.

C++ 
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

// Function to return gcd of a and b
int GcdOfArray(vector<int>& arr) {
  	int res = arr[0];
    for (int i = 1; i < arr.size(); i++) {
      
      	// Find gcd(a, b) using inbuilt library function
        res = __gcd(arr[i], res);

        if (res == 1)
            return 1;
    }
    return res;
}

int main() {
    vector<int> arr = { 2, 4, 6, 8 };
    cout << GcdOfArray(arr) << "\n";
    return 0;
}
Java 
import java.util.*;
import java.math.BigInteger;

public class GFG {

    // Function to return gcd of array
    public static int GcdOfArray(int[] array) {
        BigInteger res = BigInteger.valueOf(array[0]);

        // Find gcd(a, b) using inbuilt library function
        for (int i = 1; i < array.length; i++) {
            res = res.gcd(BigInteger.valueOf(array[i]));

            if (res.intValue() == 1)
                return 1;
        }

        return res.intValue();
    }

    public static void main(String[] args) {
        int[] array = {2, 4, 6, 8};
        System.out.println(GcdOfArray(array));
    }
}
Python 
from math import gcd

# Function to return gcd of a and b
def GcdOfArray(array):
    res = array[0]
    for num in array[1:]:
      
        # Find gcd(a, b) using inbuilt library function
        res = gcd(num, res)

        if res == 1:
            return 1
    return res

if __name__ == "__main__":
    array = [2, 4, 6, 8]
    print(GcdOfArray(array))
C# 
using System;
using System.Numerics;

class GFG {

    // Function to return gcd of a and b
    public static int GcdOfArray(int[] array) {
        BigInteger res = array[0];

        // Find gcd(a, b) using inbuilt library function
        for (int i = 1; i < array.Length; i++) {
            res = BigInteger.GreatestCommonDivisor(res, array[i]);

            if (res == 1)
                return 1;
        }

        return (int)res;
    }

    public static void Main(string[] args) {
        int[] array = {2, 4, 6, 8};
        Console.WriteLine(GcdOfArray(array));
    }
}
JavaScript 
// Function to return gcd of a and b
function GcdOfArray(array) {
    let res = array[0];
    for (let i = 1; i < array.length; i++) {

        // Find gcd(a, b) using inbuilt library function
        res = gcd(array[i], res);

        if (res === 1)
            return 1;
    }
    return res;
}

// Helper function for GCD
function gcd(a, b) {
    return b === 0 ? a : gcd(b, a % b);
}

const array = [2, 4, 6, 8];
console.log(GcdOfArray(array));

Output
2

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

Comment

Explore