Question

In Python

State g(n)'s runtime complexity:

def f(n):
   if n <= 1:`
      return 1
   return 1 + f(n/2)

def g(n):
   i = 1
   while i < n:
       f(i)
       i *= 2

Answer 1

Recursive function for f is
T(n) = T(n/2) + 1
     = T(n/4) + 1 + 1
     = T(n/8) + 1 + 1 + 1
     ......
     ......
     ......
     = T(n/n) + 1 + .... + 1 + 1 + 1 [log(n) +1 terms]
     = T(1) + 1 + .... + 1 + 1 + 1 [log(n) +1 terms]
     = 10 + 1 + .... + 1 + 1 + 1 [log(n) +1 terms]
     = clog(n)
     (Where c is a constant)
     = O(logn)
So, Time complexity of function f() = O(logn)

========================================

Time complexity of g(n):
Number of values for i are log(n)
So, The while loop in g() is running for log(n) times.
Each time in the while loop it is making call to function f()

So, Total time complexity of g(n) = O(log(n) * log(n))