The trie seems like a very good idea for your constraints. A “thinking outside the box” alternative: If you can afford some probability of answering “present” for a string that is absent EDIT: if you can afford false positives, use a Bloom filter as suggested by WizardOfOdds in the comments. For k=1, a Bloom filter … Read more
Late, but better than never I thought I would add Julian Bucknalls articles to this list. But he does not have performance numbers. In my testing of his structures the list scaled well compared to locking (very low kernel usage compared to ReaderWriterLock). His blog has a series of articles on lock free structures in … Read more
Both are product types which let you build types from multiple simpler types. Some languages treat tuples as a kind of record. Definitions A tuple is an ordered group of elements, like (10, 25). A record is typically a group of named elements like { “x”: 10, “y”: 25 } where the value has two … Read more
Python does not have a built-in data structure which meets all 3 of your requirements. That said, it’s fairly trivial to implement a tree yourself. Another option would be to combine a dictionary with a list to create what is effectively a set that also maintains a list of its items: import random class ListDict(object): … Read more
Just split your userid from behind. e.g. UserID = 6435624 Path = /images/24/56/6435624 As for the backup you could use MySQL Replication and backup the slave database to avoid problems (e.g. locks) while backuping.
The pseudocode that you’ve posted will correctly balance a tree. That said, it is too inefficient to be practical – notice that you’re recursively exploring the entire tree trying to do rebalancing operations, which will make all insertions and deletions take O(n) time, eating away all the efficiency gains of having a balanced tree. The … Read more
There are many tradeoffs involved here and I don’t think that there’s a “correct” answer to this question. If you implement the stack using a linked list with a tail pointer, then the worst-case runtime to push, pop, or peek is O(1). However, each element will have some extra overhead associated with it (namely, the … Read more
The best resource is straight from the official API: The size, isEmpty, get, set, iterator, and listIterator operations run in constant time. The add operation runs in amortized constant time, that is, adding n elements requires O(n) time. All of the other operations run in linear time (roughly speaking). The constant factor is low compared … Read more
Assuming you may destroy the input trees: remove the rightmost element for the left tree, and use it to construct a new root node, whose left child is the left tree, and whose right child is the right tree: O(log n) determine and set that node’s balance factor: O(log n). In (temporary) violation of the … Read more
The answer in a nutshell is you should generally use structured arrays rather than recarrays because structured arrays are faster and the only advantage of recarrays is to allow you to write arr.x instead of arr[‘x’], which can be a convenient shortcut, but also error prone if your column names conflict with numpy methods/attributes. See … Read more