algorithm - What does O (log n) mean exactly? - Stack Overflow
Feb 22, 2010 · 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 …
Difference between O(n) and O(log(n)) - which is better and what ...
Apr 29, 2012 · For the short answer, O (log n) is better than O (n) Now what exactly is O ( log n) ? Generally, when referring to big O notation, log n refers to the base-2 logarithm, (same way ln …
Difference between O(logn) and O(nlogn) - Stack Overflow
Mar 16, 2020 · 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 of Algorithms which has O (1), O (n log n) and O (log n ...
Jan 2, 2023 · 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 …
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 …
Why is $\log (n!)$ $O (n\log n)$? - Mathematics Stack Exchange
I thought that $\\log(n!)$ would be $\\Omega(n \\log n )$, but I read somewhere that $\\log(n!) = O(n\\log n)$. Why?
notation - What is the difference between $\log^2 (n)$, $\log (n)^2 ...
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.
What would cause an algorithm to have O(log log n) complexity?
May 23, 2017 · 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 …
which is greater? O(log*n) or O(loglog n) - Stack Overflow
Nov 21, 2013 · I have a confusion between the time complexities O(log*n) and O(loglog n) and wanted to know which one is greater. Any help?
(log(n))^log(n) and n/log(n), which is faster? - Stack Overflow
Feb 9, 2016 · 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 …