Program to Find GCD or HCF of Two Numbers

Try it on GfG Practice

Given two positive integers a and b, the task is to find the GCD of the two numbers.

Note: The GCD (Greatest Common Divisor) or HCF (Highest Common Factor) of two numbers is the largest number that divides both of them. 

Examples:

Input: a = 20, b = 28
Output: 4
Explanation: The factors of 20 are 1, 2, 4, 5, 10 and 20. The factors of 28 are 1, 2, 4, 7, 14 and 28. Among these factors, 1, 2 and 4 are the common factors of both 20 and 28. The greatest among the common factors is 4.

Input: a = 60, b = 36
Output: 12
Explanation: GCD of 60 and 36 is 12.

[Approach - 1] Using Loop - O(min(a, b)) Time and O(1) Space

The idea is to find the minimum of the two numbers and find its highest factor which is also a factor of the other number.

#include <iostream>
using namespace std;

int gcd(int a, int b)
{
    // Find Minimum of a and b
    int result = min(a, b);
    while (result > 0) {
        if (a % result == 0 && b % result == 0) {
            break;
        }
        result--;
    }

    // Return gcd of a and b
    return result;
}

int main()
{
    int a = 20, b = 28;
    cout << gcd(a, b);
    return 0;
}

import java.io.*;

class GFG {

    static int gcd(int a, int b)
    {
        // Find Minimum of a and b
        int result = Math.min(a, b);
        while (result > 0) {
            if (a % result == 0 && b % result == 0) {
                break;
            }
            result--;
        }

        // Return gcd of a and b
        return result;
    }

    public static void main(String[] args)
    {
        int a = 20, b = 28;
        System.out.print(gcd(a, b));
    }
}

def gcd(a, b):

    # Find Minimum of a and b
    result = min(a, b)

    while result > 0:
        if a % result == 0 and b % result == 0:
            break
        result -= 1

    # Return gcd of a and b
    return result


if __name__ == '__main__':
    a = 20
    b = 28
    print(gcd(a, b))

using System;

class GFG {
    
    // Function to find gcd of two numbers
    static int gcd(int a, int b) {
        // Find Minimum of a and b
        int result = Math.Min(a, b);

        while (result > 0)
        {
            if (a % result == 0 && b % result == 0)
                break;
            result--;
        }

        // Return gcd of a and b
        return result;
    }

    static void Main()
    {
        int a = 20;
        int b = 28;
        Console.WriteLine(gcd(a, b));
    }
}

function gcd(a, b) {

    // Find Minimum of a and b
    let result = Math.min(a, b);

    while (result > 0) {
        if (a % result === 0 && b % result === 0) {
            break;
        }
        result--;
    }

    // Return gcd of a and b
    return result;
}

// Driver Code
let a = 20;
let b = 28;
console.log(gcd(a, b));

4

Below both approaches are optimized approaches of the above code.

[Approach - 2] Euclidean Algorithm using Subtraction - O(min(a,b)) Time and O(min(a,b)) Space

The idea of this algorithm is, the GCD of two numbers doesn't change if the smaller number is subtracted from the bigger number. This is the Euclidean algorithm by subtraction. It is a process of repeat subtraction, carrying the result forward each time until the result is equal to any one number being subtracted.

Pseudo-code:

gcd(a, b):
    if a = b:
        return a
    if a > b:
        return gcd(a - b, b)
    else:
        return gcd(a, b - a)

#include <iostream>
using namespace std;

int gcd(int a, int b)
{
    // Everything divides 0
    if (a == 0)
        return b;
    if (b == 0)
        return a;

    // Base case
    if (a == b)
        return a;

    // a is greater
    if (a > b)
        return gcd(a - b, b);
    return gcd(a, b - a);
}

int main()
{
    int a = 20, b = 28;
    cout << gcd(a, b);
    return 0;
}

class GfG {

    static int gcd(int a, int b) {
        // Everything divides 0
        if (a == 0)
            return b;
        if (b == 0)
            return a;

        // Base case
        if (a == b)
            return a;

        // a is greater
        if (a > b)
            return gcd(a - b, b);
        return gcd(a, b - a);
    }

    public static void main(String[] args) {
        int a = 20, b = 28;
        System.out.println(gcd(a, b));
    }
}

def gcd(a, b):
  
    # Everything divides 0
    if a == 0:
        return b
    if b == 0:
        return a

    # Base case
    if a == b:
        return a

    # a is greater
    if a > b:
        return gcd(a - b, b)
    return gcd(a, b - a)

if __name__ == '__main__':
    a = 20
    b = 28
    print(gcd(a, b))

using System;

class GfG {
    static int gcd(int a, int b) {
        // Everything divides 0
        if (a == 0)
            return b;
        if (b == 0)
            return a;

        // Base case
        if (a == b)
            return a;

        // a is greater
        if (a > b)
            return gcd(a - b, b);
        return gcd(a, b - a);
    }

    static void Main()
    {
        int a = 20, b = 28;
        Console.WriteLine(gcd(a, b));
    }
}

function gcd(a, b) {
    // Everything divides 0
    if (a === 0)
        return b;
    if (b === 0)
        return a;

    // Base case
    if (a === b)
        return a;

    // a is greater
    if (a > b)
        return gcd(a - b, b);
    return gcd(a, b - a);
}

// Driver code
let a = 20, b = 28;
console.log(gcd(a, b));

4

[Approach - 3 ] Modified Euclidean Algorithm using Subtraction by Checking Divisibility - O(min(a, b)) Time and O(min(a, b)) Space

The above method can be optimized based on the following idea:

If we notice the previous approach, we can see at some point, one number becomes a factor of the other so instead of repeatedly subtracting till both become equal, we can check if it is a factor of the other.

Illustration:

See the below illustration for a better understanding:

Consider a = 98 and b = 56

a = 98, b = 56:

• a > b so put a = a-b and b remains same. So  a = 98-56 = 42  & b= 56. 

a = 42, b = 56:

• Since b > a, we check if b%a=0. Since answer is no, we proceed further. 
• Now b>a. So b = b-a and a remains same. So b = 56-42 = 14 & a= 42. 

a = 42, b = 14:

• Since a>b, we check if a%b=0. Now the answer is yes. 
• So we print smaller among a and b as H.C.F . i.e. 42 is  3 times of 14.

So HCF is 14. 

#include <iostream>
using namespace std;

int gcd(int a, int b)
{
    // Everything divides 0
    if (a == 0)
        return b;
    if (b == 0)
        return a;

    // Base case
    if (a == b)
        return a;

    // a is greater
    if (a > b) {
        if (a % b == 0)
            return b;
        return gcd(a - b, b);
    }

    // b is greater
    if (b % a == 0)
        return a;
    return gcd(a, b - a);
}

// Driver code
int main()
{
    int a = 20, b = 28;
    cout << gcd(a, b);
    return 0;
}

class GfG {

    static int gcd(int a, int b) {
        // Everything divides 0
        if (a == 0)
            return b;
        if (b == 0)
            return a;

        // Base case
        if (a == b)
            return a;

        // a is greater
        if (a > b) {
            if (a % b == 0)
                return b;
            return gcd(a - b, b);
        }

        // b is greater
        if (b % a == 0)
            return a;
        return gcd(a, b - a);
    }

    // Driver code
    public static void main(String[] args) {
        int a = 20, b = 28;
        System.out.println(gcd(a, b));
    }
}

def gcd(a, b):
    # Everything divides 0
    if a == 0:
        return b
    if b == 0:
        return a

    # Base case
    if a == b:
        return a

    # a is greater
    if a > b:
        if a % b == 0:
            return b
        return gcd(a - b, b)

    # b is greater
    if b % a == 0:
        return a
    return gcd(a, b - a)

# Driver code
if __name__ == '__main__':
    a = 20
    b = 28
    print(gcd(a, b))

using System;

class GfG
{
    static int gcd(int a, int b) {
        // Everything divides 0
        if (a == 0)
            return b;
        if (b == 0)
            return a;

        // Base case
        if (a == b)
            return a;

        // a is greater
        if (a > b)
        {
            if (a % b == 0)
                return b;
            return gcd(a - b, b);
        }

        // b is greater
        if (b % a == 0)
            return a;
        return gcd(a, b - a);
    }

    // Driver code
    static void Main()
    {
        int a = 20, b = 28;
        Console.WriteLine(gcd(a, b));
    }
}

function gcd(a, b) {
    // Everything divides 0
    if (a === 0)
        return b;
    if (b === 0)
        return a;

    // Base case
    if (a === b)
        return a;

    // a is greater
    if (a > b) {
        if (a % b === 0)
            return b;
        return gcd(a - b, b);
    }

    // b is greater
    if (b % a === 0)
        return a;
    return gcd(a, b - a);
}

// Driver code
let a = 20, b = 28;
console.log(gcd(a, b));

4

[Approach - 4] Optimized Euclidean Algorithm by Checking Remainder

Instead of the Euclidean algorithm by subtraction, a better approach can be used. We don't perform subtraction here. we continuously divide the bigger number by the smaller number. More can be learned about this efficient solution by using the modulo operator in Euclidean algorithm.

#include <iostream>
using namespace std;

// Recursive function to calculate GCD using Euclidean algorithm
int gcd(int a, int b) {
    return b == 0 ? a : gcd(b, a % b);
}

// Driver code
int main() {
    int a = 20, b = 28;
    cout << gcd(a, b); 
    return 0;
}

class GfG {

    // Recursive function to calculate GCD using Euclidean algorithm
    static int gcd(int a, int b) {
        return (b == 0) ? a : gcd(b, a % b);
    }

    // Driver code
    public static void main(String[] args) {
        int a = 20, b = 28;
        System.out.println(gcd(a, b)); 
    }
}

# Recursive function to calculate GCD using Euclidean algorithm
def gcd(a, b):
    return a if b == 0 else gcd(b, a % b)

# Driver code
a = 20
b = 28
print(gcd(a, b))  # Output: 4

using System;

class GfG
{
    // Recursive function to calculate GCD using Euclidean algorithm
    static int gcd(int a, int b) => b == 0 ? a : gcd(b, a % b);

    // Driver code
    static void Main()
    {
        int a = 20, b = 28;
        Console.WriteLine(gcd(a, b)); 
    }
}

// Recursive function to calculate GCD using Euclidean algorithm
function gcd(a, b) {
    return b === 0 ? a : gcd(b, a % b);
}

// Driver code
let a = 20, b = 28;
console.log(gcd(a, b));

4

Time Complexity: O(log(min(a,b)))

• Each recursive call reduces the size of the numbers significantly using the modulo operation (a % b), which shrinks the input faster than subtraction.
• The worst-case scenario for the number of steps occurs when the inputs are consecutive Fibonacci numbers, like (21, 13), which maximizes the number of recursive calls.
• Since Fibonacci numbers grow exponentially, and the number of steps increases linearly with their position, the time complexity becomes logarithmic in terms of the smaller number — O(log(min(a, b))).

Auxiliary Space: O(log(min(a,b))

• The maximum number of recursive calls is proportional to the number of steps taken to reduce the input to zero, which is O(log(min(a, b))) in the worst case.

[Approach - 5] Using Built-in Function - O(log(min(a, b))) Time and O(1) Space

Languages like C++ have inbuilt functions to calculate GCD of two numbers.

Below is the implementation using inbuilt functions.

#include <algorithm>
#include <iostream>
using namespace std;

int gcd(int a, int b)
{
    return __gcd(a, b);
}

// Driver code
int main()
{
    int a = 20, b = 28;
    cout << gcd(a, b);
    return 0;
}

class GfG {

    static int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }

    // Driver code
    public static void main(String[] args) {
        int a = 20, b = 28;
        System.out.println(gcd(a, b)); 
    }
}

import math

def gcd(a, b):
    return math.gcd(a, b)

# Driver code
if __name__ == '__main__':
    a = 20
    b = 28
    print(gcd(a, b))  

using System;

class GfG
{
    static int GCD(int a, int b)
    {
        return b == 0 ? a : GCD(b, a % b);
    }

    // Driver code
    static void Main()
    {
        int a = 20, b = 28;
        Console.WriteLine(GCD(a, b)); 
    }
}

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

// Driver code
let a = 20, b = 28;
console.log(gcd(a, b));

4

Please refer GCD of more than two (or array) numbers to find HCF of more than two numbers.