Find the time complexity of the following Code : int gcd(int a,int b)
{ if(b==0) return a; else Return gcd(b,a%b); }
Doubt in question 9
Hey @mansi25 ,
The time complexity for the above algorithm is O(log(min(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(min(a,b))!
It’s okay if you don’t properly understand the derivation of time complexity right now . You can always come back later and have a look at it
Hope it Helps 
. Happy Coding