Hash tables are O(1) average and amortized case complexity, however it suffers from O(n) worst case time complexity. [And I think this is where your confusion is] Hash tables suffer from O(n) worst time complexity due to two reasons: If too many elements were hashed into the same key: looking inside this key may take O(n) time. Once a hash table has passed its load balance – it has … Read more
Data in an array can be rearranged into a heap, in place. The algorithm for this is actually surprisingly simple., but I won’t go into it here. For a heap sort, you arrange the data so that it forms a heap in place, with the smallest element at the back (std::make_heap). Then you swap the … Read more
In short: Hamiltonian cycle : O(N!) in the worst case WordBreak and StringSegment : O(2^N) NQueens : O(N!) Note: For WordBreak there is an O(N^2) dynamic programming solution. More details: In Hamiltonian cycle, in each recursive call one of the remaining vertices is selected in the worst case. In each recursive call the branch factor decreases by 1. Recursion … Read more
Both have the same complexity: Constant. From the N4431 working draft, ยง21.4.4 size_type size() const noexcept; Returns: A count of the number of char-like objects currently in the string. Complexity: Constant time. And size_type length() const noexcept; Returns: size(). […] iterates through all the characters and counts the length […] That’s C strings you’re thinking … Read more
I can see how, when looking up a value in a BST we leave half the tree everytime we compare a node with the value we are looking for. However I fail to see why the time complexity is O(log(n)). So, my question is: If we have a tree of N elements, why the time … Read more
a) Yes – in a perfect world you’d have to do log n merges of size n, n/2, n/4 … (or better said 1, 2, 3 … n/4, n/2, n – they can’t be parallelized), which gives O(n). It still is O(n log n). In not-so-perfect-world you don’t have infinite number of processors and context-switching … Read more
They are not equivalent: sqrt(N) will increase a lot more quickly than log2(N). There is no constant C so that you would have sqrt(N) < C.log(N) for all values of N greater than some minimum value. An easy way to grab this, is that log2(N) will be a value close to the number of (binary) digits of N, while sqrt(N) will be a number that has itself half the … Read more
What is the complexity of creating a trie of a list of words and what is complexity of searching other set of word in that trie? Should I use trie for string searching, when i have hashtable?
For selection sort you have to search the minimum and put it on the first place in the first iteration. In the second iteration you have to search the minimum in the non-sorted part of the array and put it to the second place and so on… You only know which element is the minimum … Read more
The main purpose of the Big-O notation is to let you do the estimates like the ones you did in your post, and decide for yourself if spending your effort coding a typically more advanced algorithm is worth the additional CPU cycles that you are going to buy with that code improvement. Depending on the … Read more