Python's list implementation uses a dynamically resized C array 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.
Python's list implementation uses a dynamically resized C array 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.
To add to Martijn's answer, if you want a datastructure that has constant time pops at both ends, look at collections.deque.
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If we have a deque as : queue=collections.deque([0]) and list as : arr=[0].
Will queue.popleft() be faster ? Will arr.pop(0) be slower? Is there a difference in time complexity?
Yes. list.pop(0) is O(n), and deque.popleft() is O(1).
popleft is just a shortcut to pop(0), same way pop() is a shortcut to pop(len(sequence)-1), it's not suddenly performing a different operation with a different time complexity, as is also mentioned in the documentation
Indexed access is O(1) at both ends but slows to O(n) in the middle. For fast random access, use lists instead.
Yes, it is O(1) to pop the last element of a Python list, and O(N) to pop an arbitrary element (since the whole rest of the list has to be shifted).
Here's a great article on how Python lists are stored and manipulated: An Introduction to Python Lists.
Pop() for the last element ought to be O(1) since you only need to return the element referred to by the last element in the array and update the index of the last element. I would expect pop() for an arbitrary element to be O(N) and require on average N/2 operations since you would need to move any elements beyond the element you are removing one position up in the array of pointers.