Question

def fd(n):
    x,y,count1,count2 = n,1,0,0
    while (x > 1): (x,count1) = (x/5,1+count1)
    while (y < n): (y,count2) = (count1+y,1+count2)
    return count2

Answer 1

Let’s see. The first loop counts how many times n can be divided by 5, so count1 is log(n) (constants don’t count for O() calculations). Then the second loop counts how many times count1 (= log(n)) can be added to 1 to get to n, so it basically loops for n / log(n) times.

It looks to me like this is O(log(n) + (n/log(n))).

As J.F. Sebastian points out, n/log(n) dominates log(n), so the final answer should be O(n/log(n)).