I was wondering if someone could please explain what the time complexity is for the code below.

I think it would be $O(n)$ because the algorithm will take as much time to execute as there are elements in $n$.

sample_list = [0, 1, 2, 3, 4, 5, 5, 6]                                          
check_list = []                                                                 
for i in sample_list:                                                           
    if i not in check_list:                                                       
  • $\begingroup$ Is anything known about the implementation of check_list/[], of check_list.append()? What about i in check_list? $\endgroup$
    – greybeard
    Sep 21 at 4:52

The code takes $O(n^2)$. It is tricky to see that, since its very easy to miss the fact that if i not in sample_list takes another $O(n)$ time - just by itself.

Here is the breakdown of the complexity:

check_list = [] # O(1)
for i in sample_list # repeats n times
    if i not in check_list: # takes an O(n) time to do the check
        check_list.append(i) # takes constant time
        print(i) # again, constant time

Overall, it will take $O(n)$ per call to if i not in check_list, which will happen exactly $n$ times - and hence the total complexity is $n\cdot O(n)=O(n^2)$

  • $\begingroup$ Note that $n$ is the size of the input array here. As it stands, the given code just takes constant time. $\endgroup$
    – vonbrand
    Sep 20 at 22:36
  • 1
    $\begingroup$ I assumed sample_list is a general input, and sample_list=[0,1,2,3,4,5,6] is just an example. You are right that technically the code as it is right now takes $O(1)$. $\endgroup$
    – nir shahar
    Sep 20 at 23:00
  • $\begingroup$ I would have answered like @nirshahar, I would assume the array is not constant, it is just an example to run the code. The question seems to me about the algorithm, not the a specific instance. Any ready to run code can be seen as $O(1)$ if we consider every single array we see in an example a constant... $\endgroup$
    – user206904
    Oct 20 at 21:54

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