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…)
MSDN Lists these: Dictionary<,> List<> SortedList<,> (edit: wrong link; here’s the generic version) SortedDictionary<,> etc. For example: The SortedList(TKey, TValue) generic class is a binary search tree with O(log n) retrieval, where n is the number of elements in the dictionary. In this, it is similar to the SortedDictionary(TKey, TValue) generic class. The two classes … Read more
Theorem: Every deterministic algorithm for this problem probes Ω(log2 n) memory locations in the worst case. Proof (completely rewritten in a more formal style): Let k > 0 be an odd integer and let n = k2. We describe an adversary that forces (log2 (k + 1))2 = Ω(log2 n) probes. We call the maximal … Read more
During every iteration you increment i and decrement j which is equivalent to just incrementing i by 2. Therefore, total number of iterations is n^2 / 2 and that is still O(n^2).
Since you have a set of unique values, a Set is the appropriate data structure. You can put your values inside HashSet, an implementation of the Set interface. My lead said this was a more efficient way to do this and it made sense to me, but it just seemed a bit off to put … Read more
According to a recent episode of Radiolab (about “Falling”), a cat reaches terminal velocity by the 9th floor. After that, it relaxes and is less likely to be hurt. There are completely uninjured cats after a fall from above the 30th. The riskiest floors are 5th to 9th.
f ∈ O(g) says, essentially For at least one choice of a constant k > 0, you can find a constant a such that the inequality 0 <= f(x) <= k g(x) holds for all x > a. Note that O(g) is the set of all functions for which this condition holds. f ∈ o(g) … Read more