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I know that the time complexity of quick sort in the worst case is $\Theta(n^2)$ and in the average case is $\Theta(n \log n)$. Can it be $\Theta(n\sqrt n)$ for certain inputs?

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    $\begingroup$ In the average case it's $\Theta(n\log n)$. It can probably be $\Theta(n\sqrt{n})$ for certain inputs of a certain structure. I suggest looking up the best and worst cases for quicksort, and trying to combine them. $\endgroup$ – Yuval Filmus Feb 7 '18 at 13:08
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Assume the array is sorted. If you always pick the element at n/2 as pivot then the time is O(n log n). If you always pick the first or last element it’s $O(n^2)$.

So which pivot gives you $O(n^{1.5})$?

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  • $\begingroup$ The third quarter? $\endgroup$ – Ahmad Feb 8 '18 at 5:54
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    $\begingroup$ Can you give a formal proof? $\endgroup$ – Mr. Sigma. Feb 8 '18 at 12:37
  • $\begingroup$ This answer is very much incomplete. And misleading too, for any constant $c$, if you always pick the element at $cn$ as your pivot the complexity is $\Theta(n \log n)$, and never $\Theta(n\sqrt n)$. $\endgroup$ – orlp Sep 28 at 8:17

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