Dec 2, 2018 · How come O (n) + O (logn) = O (logn)? When talking for example about an algorithm that has two operations. One of them takes O (n) and the other O (logn) and in the end we say that the total complexity is O (logn). It doesn't make …

Similarly even if an algorithm has O(log n) complexity, log n cannot possibly grow larger than about a hundred, so could be ignored as no larger than a small constant. So, is it really meaningful to treat O(1) and O(log n) as different? The same applies of any difference of log n, like between O(n), O(n log n) and O(n/ log n).

Oct 17, 2015 · I tried: $\log (n!) = \log1 + \dots + \log n \leq n \log n \Rightarrow \log (n!) = O (n \log n)$. But how can we prove $\log (n!) = \Omega (n \log n)$ without Sterli...

Apr 18, 2022 · This sort of got me thinking, what are the conditions under which you can do a O (log n) search for an element in an array in which all numbers are distinct? Clearly, you can still do a O (log n) search even if the array is something less than strictly sorted.

OK but we already have two answers saying "change variables to cm c m, solve that recurrence and substitute to get T(n) = Θ(log n log log n) T (n) = Θ (log n log log n). So what does your answer add? (And note that your form of the answer confusingly uses different bases in the logarithms: the suppressed constant factors in the big-Theta notation let you change all the bases to be the same.)

By definition, it means that there exists c,n0 c, n 0 such that for every n ≥ n0 n ≥ n 0, n2 <c log n n 2 <c log n, that is, n2 log n <c n 2 log n <c. Now you need to explain why this is absurd.

Had a question about the following: $$\\log (n+1) \\in O(\\log n)$$ Can the left side be simplified any further or do I need to just go ahead and find a c and n that hold?

Jan 8, 2016 · Log^2 (n) means that it's proportional to the log of the log for a problem of size n. Log (n)^2 means that it's proportional to the square of the log.

Your archetypical Θ(n log n) Θ (n log n) is a divide-and-conquer algorithm, which divides (and recombines) the work in linear time and recurses over the pieces. Merge sort works that way: spend O(n) O (n) time splitting the input into two roughly equal pieces, recursively sort each piece, and spend Θ(n) Θ (n) time combining the two sorted halves. Intuitively, continuing the divide-and ...

A large part of algorithmic complexity is to determine which algorithm is eventually faster, thus knowing that O(n^f) is faster than O(n/log n) for 0 < f < 1, is often enough.