Hello Everyone,
GCD (Greatest Common Divisor) or HCF (Highest Common Factor) of two numbers is the largest number that divides both of them.
An efficient solution is to use Euclidean algorithm which is the main algorithm used for this purpose. The idea is, GCD of two numbers doesn’t change if smaller number is subtracted from a bigger number.
// C++ program to find GCD of two numbers
#include <iostream>
using namespace std;
// Recursive function to return gcd of a and b
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);
}
// Driver program to test above function
int main()
{
int a = 98, b = 56;
cout<< "GCD of " <<a<< " and " <<b<< " is " <<gcd(a, b);
return 0;
}
Output:
GCD of 98 and 56 is 14
// C++ program to find GCD of two numbers
#include <iostream>
using namespace std;
// Recursive function to return gcd of a and b
int gcd( int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Driver program to test above function
int main()
{
int a = 98, b = 56;
cout<< "GCD of " <<a<< " and " <<b<< " is " <<gcd(a, b);
return 0;
}
Output:
GCD of 98 and 56 is 14
The time complexity for the above algorithm is O(log(max(a,b))) the derivation for this is obtained from the analysis of the worst-case scenario. What we do is we ask what are the 2 least numbers that take 1 step, those would be (1,1). If we want to increase the number of steps to 2 while keeping the numbers as low as possible as we can take the numbers to be (1,2). Similarly, for 3 steps, the numbers would be (2,3), 4 would be (3,5), 5 would be (5,8). So we can notice a pattern here, for the nth step the numbers would be (fib(n),fib(n+1)). So the worst-case time complexity would be O(n) where a>= fib(n) and b>= fib(n+1).
Now Fibonacci series is an exponentially growing series where the ratio of nth/(n-1)th term approaches (sqrt(5)-1)/2 which is also called the golden ratio. So we can see that the time complexity of the algorithm increases linearly as the terms grow exponentially hence the time complexity would be log(max(a,b)).