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algorithm - What does O (log n) mean exactly? - Stack Overflow You can think of O (1), O (n), O (logn), etc as classes or categories of growth Some categories will take more time to do than others These categories help give us a way of ordering the algorithm performance Some grown faster as the input n grows The following table demonstrates said growth numerically
Difference between O(logn) and O(nlogn) - Stack Overflow I am preparing for software development interviews, I always faced the problem in distinguishing the difference between O(logn) and O(nLogn) Can anyone explain me with some examples or share some
Examples of Algorithms which has O (1), O (n log n) and O (log n . . . O (1) - most cooking procedures are O (1), that is, it takes a constant amount of time even if there are more people to cook for (to a degree, because you could run out of space in your pot pans and need to split up the cooking) O (logn) - finding something in your telephone book Think binary search O (n) - reading a book, where n is the number of pages It is the minimum amount of time it
algorithm - Is log (n!) = Θ (n·log (n))? - Stack Overflow @Z3d4s the what steps 7-8 conversion is saying that n logn == log (n^n) and for showing the bound here you can say the first term is always greater than the second term you can check for any larger values, and for expressing big-O complexity we will always take the dominating item of all So n logn contributes to the big-O time
What is O (log (n!)), O (n!), and Stirlings approximation? By Stirling's approximation, log(n!) = n log(n) - n + O(log(n)) For large n, the right side is dominated by the term n log (n) That implies that O (log (n!)) = O (n log (n)) More formally, one definition of "Big O" is that f (x) = O (g (x)) if and only if lim sup|f(x) g(x)| < ∞ as x → ∞ Using Stirling's approximation, it's easy to show that log (n!) ∈ O (n log (n)) using this
What would cause an algorithm to have O(log log n) complexity? 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 algorithm to work by repeatedly cut the size of the input down by some constant factor on each
(log(n))^log(n) and n log(n), which is faster? - Stack Overflow Short answer: Yes Longer answer, run a profiler on the code big-o is not usable for actual performance measurements, only for "what happens if N grows towards infinity" type of problems And what does "O (x)" have to do with the problem? if f=O(g(n)), is n a constant? If not, why is it not f(n)=O(g(n))? Or is f related to f(n)=(log(n))^log(n)?