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

def f(n): c = 0 for i in range(n): c = c + 1 for j in range(n): c = c + 1 for k in range(n): c = c + 1 return c 

O(3^n)

O(n)

O(n^3)

O(n^2)

def f(n): c = 0 for i in range(n): s = n while s > 1: s = s/2 c = c + 1 return c 

O(n*log(n))

O(n)

O(n^2)

O(n*2^n)