Do I understand n log n correctly?

vaderboi

Junior Member
Joined
Jan 18, 2019
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59
I just want to make sure I understand "n log n" correctly. So "n log n" is equivalent to "log n^n", correct? And "log n^n" is equivalent to "base^y = n^n"?

Also, in terms of big-Oh notation: when it is said that O(n log n) has a more expensive worst case time complexity than O(n), does that remain true for any base in O(n log n)?
 
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