2010年2月22日 · O(log N) basically means time goes up linearly while the n goes up exponentially. So if it takes 1 second to compute 10 elements, it will take 2 seconds to compute 100 elements, 3 seconds to compute 1000 elements, and so on. It is O(log n) when we do divide and conquer type of algorithms e

2010年3月5日 · The log* N bit is an iterated algorithm which grows very slowly, much slower than just log N.You basically just keep iteratively 'logging' the answer until it gets below one (E.g: log(log(log(...log(N)))), and the number of times you had to log() is the answer.

2022年7月19日 · 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:

2012年4月29日 · Whereas, O(log n) means when input size 'n' increases exponentially, our running time will increase linearly. Note that it might happen that O(log n) is faster than O(1) in some cases but O(1) will outperform O(log n) when n grows as O(1) is independent of the input size n. Considering these two code snippets,

2019年4月27日 · Think of it as O(n*log(n)), i.e. "doing log(n) work n times". For example, searching for an element in a sorted list of length n is O(log(n)). Searching for the element in n different sorted lists, each of length n is O(n*log(n)). Remember that O(n) is defined relative to some real quantity n. This might be the size of a list, or the number of ...

2020年7月9日 · O(n*log(n)) is clearly greater than O(n) for n>2 (log base 2) An easy way to remember might be, taking two examples. Imagine the binary search algorithm with is Log N time complexity: O(log(N)) If, for each step of binary search, you had to iterate the array of N elements; The time complexity of that task would be O(N*log(N))

2009年10月20日 · A typical example of O(N log N) would be sorting an input array with a good algorithm (e.g. mergesort). A typical example if O(log N) would be looking up a value in a sorted input array by bisection.

2017年5月23日 · Since n = 2 k, this means that k = log 2 n, and therefore the number of square roots taken is O(log k) = O(log log n). Therefore, if there is algorithm that works by repeatedly reducing the problem to a subproblem of size that is the square root of the original problem size, that algorithm will terminate after O(log log n) steps.

2013年11月21日 · O(log*N) < O(loglogN) for sufficiently large input. log*n is defined as taking log of number till it amounts to 1. The inverse function of log*n is a tower of 2 to power of 2's which increases extremely fast hence log*n grows very slowly.

Another thing that you need to keep in mind is the space complexity. Very often, algorithms with O(N*Log N) time complexity would have an O(Log N) space complexity. This may present a problem for extremely large data sets, for example when a recursive function runs on a system with a limited stack capacity.