The time complexity is O(N) on average, O(NM) worst case (N being the length of the longer string, M, the shorter string you search for). As of Python 3.10, heuristics are used to lower the worst-case scenario to O(N + M) by switching algorithms. The same algorithm is used for str.index(), str.find(), str.__contains__() (the in … Read more
As alluded to in the comments by btilly and Amadan, this is due to how the Timsort sorting algorithm works. Detailed description of the algorithm is here. Timsort speeds up operation on partially sorted arrays by identifying runs of sorted elements. A run is either “ascending”, which means non-decreasing: a0 <= a1 <= a2 <= … Read more
Kruskal is O(E log E); your derivation is right. You could also say O(E log V) because E <= V * V, so log(E) <= 2 log(V) (I don’t know why I remember that, other than that I think a prof put that on an exam at one point…)
The time complexities for BFS and DFS are just O(|E|), or in your case, O(m). In a binary tree, m is equal to n-1 so the time complexity is equivalent to O(|V|). m refers to the total number of edges, not the average number of adjacent edges per vertex.
A simple example of algorithm that is O(m+n): int sum(int[] nArr, int[] mArr) { int sum = 0; for(int i : nArr) { sum += i; } for(int i : mArr) { sum += i; } return sum; } To compute the sum, you need to go through all elements in nArr (size n) and … Read more
Yes, you are right, it is O(n) where n – length of list. Look here for more information: https://www.ics.uci.edu/~pattis/ICS-33/lectures/complexitypython.txt
The famous mathematician Gauss is said to have found a formula for that exact problem when he was in primary school. And as mentioned by @Henry in the comments it is: Source: Wikipedia (DE), Wikipedia (EN) As work is done for every entry, i.e., O(1) is required for each “item”. Hence, the problem is in … Read more
Proving an upper bound means you have proven that the algorithm will use no more than some limit on a resource. Proving a lower bound means you have proven that the algorithm will use no less than some limit on a resource. “Resource” in this context could be time, memory, bandwidth, or something else.
Here is a near O(1) solution for a Fibonacci sequence term. Admittedly, O(log n) depending on the system Math.pow() implementation, but it is Fibonacci w/o a visible loop, if your interviewer is looking for that. The ceil() was due to rounding precision on larger values returning .9 repeating. Example in JS: function fib (n) { … Read more
This is because the article that you are reading considered “getting to that index” as a separate operation. The article assumes that you are already at the index you wish to perform add(int, E). To conclude: Insert or Remove operation = O(1) Finding node at nth index = O(n)