This question already has an answer here:
Shouldn't they both be O(1)
, as popping an element from any location in a Python list involves destroying that list and creating one at a new memory location?
This question already has an answer here:
Shouldn't they both be O(1)
, as popping an element from any location in a Python list involves destroying that list and creating one at a new memory location?
To add to Martijn's answer, if you want a datastructure that has constant time pops at both ends, look at
collections.deque
.Python's
list
implementation uses a dynamically resized Carray
under the hood, removing elements usually requires you to move elements following after up to prevent gaps.list.pop()
with no arguments removes the last element. Accessing that element can be done in constant time. There are no elements following so nothing needs to be shifted.list.pop(0)
removes the first element. All remaining elements have to be shifted up one step, so that takes O(n) linear time.