In Python, a list has list.insert(i, x)
to "Insert an item at a given position.". In C++, there is a list as well. In C++, cost/complexity of inserting an element anywhere is O(1). Is it the same for a Python list? If not, can anything else be use to get O(1) insert time in Python?
Python List insert() Time Complexity. The time complexity to insert an element at a given index is O(n). This means that the complexity increases linearly with the number of elements in the list.
If you don't know the location, then you need to traverse the list to the location of deletion/insertion, which takes O(n) time. ** The deletion cost is O(log n) for the minimum or maximum, O(n) for an arbitrary element. There is some confusion in deletion in array.
Worst time complexity for inserting into list is O(n-i) , where n is the length of the list and i is the index at which you are inserting. So in case if you are inserting at the last index, that makes it O(1) .
Time Complexity: Append has constant time complexity i.e.,O(1). Extend has a time complexity of O(k). Where k is the length of the list which need to be added.
list
The Average Case assumes parameters generated uniformly at random.
Internally, a list is represented as an array; the largest costs come from growing beyond the current allocation size (because everything must move), or from inserting or deleting somewhere near the beginning (because everything after that must move). If you need to add/remove at both ends, consider using a collections.deque instead.
So inserting an element at given position will always have the time complexity of O(n) as both insert method and slicing has time complexity of O(n) and O(k). Only append which inserts at the end of list have O(1) time complexity. From Python Wiki
Lists: Complexity Operation | Example | Class | Notes --------------+--------------+---------------+------------------------------- Index | l[i] | O(1) | Store | l[i] = 0 | O(1) | Length | len(l) | O(1) | Append | l.append(5) | O(1) | Clear | l.clear() | O(1) | similar to l = [] Slice | l[a:b] | O(b-a) | l[1:5]:O(l)/l[:]:O(len(l)-0)=O(N) Extend | l.extend(...)| O(len(...)) | depends only on len of extension Construction | list(...) | len(...) | depends on lenghth of argument check ==, != | l1 == l2 | O(N) | Insert | l[a:b] = ... | O(N) | Delete | del l[i] | O(N) | Remove | l.remove(...)| O(N) | Containment | x in/not in l| O(N) | searches list Copy | l.copy() | O(N) | Same as l[:] which is O(N) Pop | l.pop(...) | O(N) | Pop | l.pop() | O(1) | same as l.pop(-1), popping at end Extreme value | min(l)/max(l)| O(N) | Reverse | l.reverse() | O(N) | Iteration | for v in l: | O(N) | Sort | l.sort() | O(N Log N) | key/reverse doesn't change this Multiply | k*l | O(k N) | 5*l is O(N): len(l)*l is O(N**2)
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