This follows easily from a simple fact in Calculus: and we have the following inequality: Here we can conclude that S = 1 + 1/2 + … + 1/n is both Ω(log(n)) and O(log(n)), thus it is Ɵ(log(n)), the bound is actually tight.
To understand the difference between polynomial time and pseudopolynomial time, we need to start off by formalizing what “polynomial time” means. The common intuition for polynomial time is “time O(nk) for some k.” For example, selection sort runs in time O(n2), which is polynomial time, while brute-force solving TSP takes time O(n · n!), which … Read more
O(n!) isn’t equivalent to O(n^n). It is asymptotically less than O(n^n). O(log(n!)) is equal to O(n log(n)). Here is one way to prove that: Note that by using the log rule log(mn) = log(m) + log(n) we can see that: log(n!) = log(n*(n-1)*…2*1) = log(n) + log(n-1) + … log(2) + log(1) Proof that O(log(n!)) … Read more
BTW complexity is indeed O(n^3) to lower that you need to: sort the points somehow by x and or y in ascending or descending order. Also use of polar coordinates can help sometimes use divide at impera (divide and conquer) algorithms usually for planar geometry algorithms is good idea to divide area to quadrants and … Read more
O(log log n) terms can show up in a variety of different places, but there are typically two main routes that will arrive at this runtime. Shrinking by a Square Root As mentioned in the answer to the linked question, a common way for an algorithm to have time complexity O(log n) is for that … Read more
One trick for analyzing the time complexity of Euclid’s algorithm is to follow what happens over two iterations: a’, b’ := a % b, b % (a % b) Now a and b will both decrease, instead of only one, which makes the analysis easier. You can divide it into cases: Tiny A: 2a <= … Read more
O(n!) isn’t equivalent to O(n^n). It is asymptotically less than O(n^n). O(log(n!)) is equal to O(n log(n)). Here is one way to prove that: Note that by using the log rule log(mn) = log(m) + log(n) we can see that: log(n!) = log(n*(n-1)*…2*1) = log(n) + log(n-1) + … log(2) + log(1) Proof that O(log(n!)) … Read more
Short answer: str slices, in general, copy. That means that your function that does a slice for each of your string’s n suffixes is doing O(n2) work. That said, you can avoid copies if you can work with bytes-like objects using memoryviews to get zero-copy views of the original bytes data. See How you can … Read more
The ECMA specification does not specify a bounding complexity, however, you can derive one from the specification’s algorithms. push is O(1), however, in practice it will encounter an O(N) copy costs at engine defined boundaries as the slot array needs to be reallocated. These boundaries are typically logarithmic. pop is O(1) with a similar caveat … Read more
The complexity of in depends entirely on what L is. e in L will become L.__contains__(e). See this time complexity document for the complexity of several built-in types. Here is the summary for in: list – Average: O(n) set/dict – Average: O(1), Worst: O(n) The O(n) worst case for sets and dicts is very uncommon, … Read more