# Find GCD of two numbers

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(logɸmin(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(logɸmin(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