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Not everything can be done in constant time, that's O(k)
(sh.itjust.works)
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this post was submitted on 22 Jun 2024
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I know I shouldn't do it here, but let me ask a serious question : does the square in O(n!²) really matter ? I have a confused intuition that the factorial grows so much faster than the square that it kind of disapears assymptoticaly.
no it's the joke. In o-notation you always use the highest approximation, so o(n!²) does not exist, it's only o(n!)
Otherwise there would never be o(1) or o(n), because o(1) would imply that the algorithm only has a single line of instructions, same for o(n)
T = O(n) means that there exists a single constant k such that T < kn for all sufficiently large n. Therefore O(n!^2) is not the the same as O(n!), but for example both 10n!, 10000n!, n! + n^2 (note the plus) are O(n!).
Another way to think about this: suppose you believe that O(n) and O(n^2) are distinct. Now plug in only numbers that are factorials (2, 6, 24, ...).