Big-O of list slicing

Question:

Say I have some Python list, my_list which contains N elements. Single elements may be indexed by using my_list[i_1], where i_1 is the index of the desired element. However, Python lists may also be indexed my_list[i_1:i_2] where a “slice” of the list from i_1 to i_2 is desired. What is the Big-O (worst-case) notation to slice a list of size N?

Personally, if I were coding the “slicer” I would iterate from i_1 to i_2, generate a new list and return it, implying O(N), is this how Python does it?

Thank you,

Asked By: mjgpy3

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Answers:

Getting a slice is O(i_2 - i_1). This is because Python’s internal representation of a list is an array, so you can start at i_1 and iterate to i_2.

For more information, see the Python Time Complexity wiki entry

You can also look at the implementation in the CPython source if you want to.

Answered By: Sam Mussmann

For a list of size N, and a slice of size M, the iteration is actually only O(M), not O(N). Since M is often << N, this makes a big difference.

In fact, if you think about your explanation, you can see why. You’re only iterating from i_1 to i_2, not from 0 to i_1, then I_1 to i_2.

Answered By: abarnert

according to http://wiki.python.org/moin/TimeComplexity

it is O(k) where k is the slice size

Answered By: Joran Beasley
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