Program to Find GCD or HCF of Two Numbers (original) (raw)

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.

gcd

**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.

Table of Content

[Approach - 1] Using Loop - O(min(a, b)) Time and O(1) Space
[Approach - 2] Euclidean Algorithm using Subtraction - O(min(a,b)) Time and O(min(a,b)) Space
[Approach - 3 ] Modified Euclidean Algorithm using Subtraction by Checking Divisibility - O(min(a, b)) Time and O(min(a, b)) Space
[Approach - 4] Optimized Euclidean Algorithm by Checking Remainder
[Approach - 5] Using Built-in Function - O(log(min(a, b))) Time and O(1) Space

[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.

C++ `

#include using namespace std;

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

Java

import java.io.*;

class GFG {

static int gcd(int a, int b)
{
    // Everything divides 0
    if(a==0 || b==0) return Math.max(a, 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));
}

}

Python

def gcd(a, b):

# Everything divides 0
if(a==0 or b==0):
    return max(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))

C#

using System;

class GFG {

// Function to find gcd of two numbers
static int gcd(int a, int b) {
    
    // Everything divides 0
    if(a==0 || b==0) return Math.Max(a, 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));
}

}

JavaScript

function gcd(a, b) {

// Everything divides 0
if(a==0 || b==0) return Math.max(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));

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)

C++ `

#include 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; }

Java

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));
}

}

Python

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))

C#

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));
}

}

JavaScript

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));

[Approach - 3 ] Modified Euclidean Algorithm using Subtraction byChecking 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.

C++ `

#include 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; }

Java

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));
}

}

Python

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))

C#

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));
}

}

JavaScript

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));

[Approach - 4]Optimized EuclideanAlgorithm 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.

C++ `

#include 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; }

Java

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)); 
}

}

Python

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

C#

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)); 
}

}

JavaScript

// 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));

**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.

C++ `

#include #include 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; }

Java

import java.math.BigInteger;

class GfG {

public static void main(String[] args) {
    int a = 20, b = 28;

    // Convert to BigInteger and compute GCD
    int gcd = BigInteger.valueOf(a).gcd(BigInteger.valueOf(b)).intValue();

    System.out.println(gcd); 
}

}

Python

import math

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

Driver code

if name == 'main': a = 20 b = 28 print(gcd(a, b))

C#

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)); 
}

}

JavaScript

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

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

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