What is the time complexity of this do-while loop?

Question

int count=0;
do
{
    count++;
    n=n/2;

}while (n>1);

I'm having trouble seeing the pattern here.
Even when plugging in numbers for n and then plotting out each basic operation.
Thanks in advance!

Edit: I want the worst case here.

In practice, it is O(1) because `INT_MAX` is a constant and the growth is so slow. — Raymond Chen, Oct 01 '17 at 23:42
See: https://stackoverflow.com/questions/9152890/what-would-cause-an-algorithm-to-have-olog-n-complexity — user2205930, Oct 02 '17 at 01:07

Adnan Isajbegovic · Answer 1 · 2017-10-02T14:17:32.783

First step is to divide n with 2. So you get n/2. Now, you divide it again with 2, if n/2 > 1, and you get n/4. If n/4 > 1 you do it again, and you get n/8, or better to write it as n/(2^3)... Now if n/(2^3) > 1 you do it again, and you get n/(2^4)... So, if you do it k times, you get n/(2^k). How to calculate k to so you get n/(2^k) ≤ 1? Easy:

n/(2^k) ≤ 1
n ≤ 2^k
ln(n) ≤ k

So, your algorithm needs O(ln(n)) iterations to exit the loop.

In your code, k is count.

What is the time complexity of this do-while loop?

1 Answers1