For example, suppose algorithm 1 requires N 2 time, and algorithm 2 requires 10 * N 2 N time For both algorithms, the time is O(N 2 ), but algorithm 1 will always be faster than algorithm 2 In this case, the constants and loworder terms do matter in terms of which algorithm is n log 2 2 n 1 = n Finally, we can see that recursion runtime from step 2) is O(n) and also the nonrecursion runtime is O(n) So, we have the case 2 O(n log b a log(n)) O(n log 2 2 log(n)) O(n 1 log(n)) O(n log(n)) 👈 this is running time of the merge sort O(2^n) Exponential timeValid, yes You can express any growth/complexity function inside the BigOh notation As others have said, the reason why you do not encounter it really often is because it looks sloppy as it is trivial to additionally show that mathn/2 \in O(n
Big O Notation Understanding Time Complexity Using Flowcharts Dev Community
