One important thing most people forget when talking about Big-O, thus I feel the need to mention that: You cannot use Big-O to compare the speed of two algorithms. Big-O only says how much slower an algorithm will get (approximately) if you double the number of items processed, or how much faster it will get … Read more
Your intuition behind why the algorithm should be Θ(n2) is a good one, but most suffix trees are designed in a way that eliminates the need for this time complexity. Intuitively, it would seem that you need Θ(n2) different nodes to hold all of the different suffixes, because you’d need n + (n – 1) … Read more
Actually, I think access based on an index will be O(log(n)), because you’ll still be searching down through a B-Tree-esque organization to get to your record.
What you have coded in your example is very similar to a depth first search. So, that’s one answer. A depth first search algorithm without any special characteristics ( like re-convergent paths that can be optimized out ), should be n^n. This is actually not a contrived example. Chess programs operate on the same algorithm. … Read more
Consider: f1(n) = n2 f2(n) = n + 1000 Clearly f1 is O(n2) and f2 is O(n). For a small input (say, n=5), we have f1(n) = 25 but f2(n) > 1000. Just because one function (or time complexity) is O(n) and another is O(n2) doesn’t mean that the former is smaller for all values … Read more
One of the best known algorithms is the Boyer-Moore string searching algorithm which is O(n) although it can give sublinear performance in the best case. Which algorithm is used in Java depends on which implemetation you download. It seems that for example OpenJDK uses a naive algorithm that runs in O(nm) and linear performance in … Read more
Most relational databases structure indices as B-trees. If a table has a clustering index, the data pages are stored as the leaf nodes of the B-tree. Essentially, the clustering index becomes the table. For tables w/o a clustering index, the data pages of the table are stored in a heap. Any non-clustered indices are B-trees … Read more
Information on this topic is now available on Wikipedia at: Search data structure +———————-+———-+————+———-+————–+ | | 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) … Read more
Think of it as O(n*log(n)), i.e. “doing log(n) work n times”. For example, searching for an element in a sorted list of length n is O(log(n)). Searching for the element in n different sorted lists, each of length n is O(n*log(n)). Remember that O(n) is defined relative to some real quantity n. This might be … Read more
In the draft C++11 standard N3337 the answer can be found in § 24.2.1 paragraph 8: All the categories of iterators require only those functions that are realizable for a given category in constant time (amortized). Since each operation on an iterator must be constant time, iterating through n elements must be O(n).