For the following blocks of code, determine their best (lowest) worst-case big-O complexity as a function of n:

def f(n):

 c = n^100 for i in range(100): c = c - n return c 

O(1)

O(100^n)

O(n^100)

O(n)

 def f(n): c = 0 step = n while step > 1: step = step/2 c = c + 1 return c 

O(n^2)

O(log(n))

O(n*log(n))

O(n)