Find GCD of two numbers - Tutorial [Updated]

Problem Statement: Find gcd of two numbers.

Examples:

Example 1:
Input: num1 = 4, num2 = 8
Output: 4
Explanation: Since 4 is the greatest number which divides both num1 and num2.

Example 2:
Input: num1 = 3, num2 = 6
Output: 3
Explanation: Since 3 is the greatest number which divides both num1 and num2.

Solution

Disclaimer: Don’t jump directly to the solution, try it out yourself first.

Solution 1: Brute force

Intuition: Simply traverse from 1 to min(a,b) and check for every number.

Approach:

Traverse from 1 to min(a,b).
And check if i is divisible by both a and b.If yes store it in the answer.
Find the maximum value of i which divides both a and b.

Code:

C++ Code

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int num1 = 4, num2 = 8;
	int ans;
	for (int i = 1; i <= min(num1, num2); i++) {
		if (num1 % i == 0 && num2 % i == 0) {
			ans = i;
		}
	}
	cout << "The GCD of the two numbers is "<<ans;
}

Output:

The GCD of the two numbers is 4

Time Complexity: O(N).

Space Complexity: O(1).

Java Code

public class Main {
  public static void main(String args[]) {
    int num1 = 3, num2 = 6;
    int ans = 1;
    for (int i = 1; i <= Math.min(num1, num2); i++) {
      if (num1 % i == 0 && num2 % i == 0) {
        ans = i;
      }
    }
    System.out.print("The GCD of the two number is "+ans);
  }
}

Output:

The GCD of the two numbers is 3

Time Complexity: O(N).

Space Complexity: O(1).

Solution 2: Using Euclidean’s theorem.

Intuition: Gcd is the greatest number which is divided by both a and b.If a number is divided by both a and b, it is should be divided by (a-b) and b as well.

Approach:

Recursively call gcd(b,a%b) function till the base condition is hit.
b==0 is the base condition.When base condition is hit return a,as gcd(a,0) is equal to a.

Code:

C++ Code

#include<bits/stdc++.h>
using namespace std;
int gcd(int a, int b) {
	if (b == 0) {
		return a;
	}
	return gcd(b, a % b);
}
int main()
{

	int a = 4, b = 8;
	cout <<"The GCD of the two numbers is "<<gcd(a, b);
}

Output:

The GCD of the two numbers is 4

Time Complexity: O(log_ɸmin(a,b)),where ɸ is 1.61.

Space Complexity: O(1).

Java Code

public class Main {
  static int gcd(int a, int b) {
    if (b == 0) {
      return a;
    }
    return gcd(b, a % b);
  }
  public static void main(String args[]) {
    int a = 4, b = 8;
    int ans = gcd(a, b);
    System.out.print("The GCD of the two numbers is "+ans);
  }
}

Output:

The GCD of the two numbers is 4

Time Complexity: O(log_ɸmin(a,b)),where ɸ is 1.61.

Space Complexity: O(1).

Special thanks to Pranav Padawe for contributing to this article on takeUforward. If you also wish to share your knowledge with the takeUforward fam, please check out this article