Isn't insertion in a (unsorted) dynamic array O(1) instead of O(n)?
pakmans opened this issue · 3 comments
pakmans commented
Isn't insertion in a (unsorted) dynamic array O(1) instead of O(n)?
amejiarosario commented
@pakmans Yes, it is!
The actual table should be like this:
+----------------------+----------+------------+----------+--------------+
| | Insert | Delete | Search | Space Usage |
+----------------------+----------+------------+----------+--------------+
| Unsorted array | O(1) | O(1) | O(n) | O(n) |
| Value-indexed array | O(1) | O(1) | O(1) | O(n) |
| Sorted array | O(n) | O(n) | O(log n) | O(n) |
| Unsorted linked list | O(1)* | O(1)* | O(n) | O(n) |
| Sorted linked list | O(n)* | O(1)* | O(n) | O(n) |
| Balanced binary tree | O(log n) | O(log n) | O(log n) | O(n) |
| Heap | O(log n) | O(log n)** | O(n) | O(n) |
| Hash table | O(1) | O(1) | O(1) | O(n) |
+----------------------+----------+------------+----------+--------------+
* The cost to add or delete an element into a known location in the list (i.e. if you have an iterator to the location) is O(1). 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.