What is the time complexity of zip() in Python?

Question

How can I calculate the time complexity of zip()?

testList = [[1,2,3]for _ in range(5)]
zip(*testList)

Answer 1

Assume you zip N iterables.

In python 3.x , the zip function itself runs in O(1) time, as it just allocates a special iterable (called the zip object), and assigns the parameter array to an internal field. The function invocation itself (before control reaches in zip) is O(N) , as the interpreter must convert the parameters to an array. Every subsequent next call on the iterator also runs in O(N) . Exhausting the zip object is therefore O(N*M) assuming M is the average (or minimum) length of the iterables, excluding the time the iterables themselves take to generate items (as it is independent of zip).

In python 2.x , the zip function returns a list. That list must be constructed during the call, that is equvivalent to exhausting the iterator in the previous example, so O(N*M) , not counting the time spent in the zipped iterables.