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Can $n$ elements be sorted in a worst case time of $\sqrt{n}\log n$? Why or why not? I've seen algorithms being sorted in the worst case of $n\log n$, so why can they be or cannot be sorted in $\sqrt{n}\log n$?

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  • $\begingroup$ Depend on the data. See counting sort $\endgroup$ – kelalaka Nov 25 '18 at 18:13
  • $\begingroup$ Without any assumptions on the data, any comparison based sorting algorithm takes $\Omega(n \log n)$. See here $\endgroup$ – Gokul Nov 25 '18 at 18:15
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    $\begingroup$ Have you consulted a textbook, or any reputable resource about it? What makes you think about $\sqrt(n)$, specifically? $\endgroup$ – Raphael Nov 25 '18 at 19:07
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This is impossible. You cannot sort $n$ elements without examining each element at least once, which takes at least $n$ operations.

$n=\sqrt{n}\cdot\sqrt{n}\ge\sqrt{n}\cdot\log{n}$ for all $n>1$, assuming the logarithmic base is at least 2. (Of course, if you assume something completely asinine like $\log_{1.000001}n$, it would be technically possible for some values of $n$.)

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