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Problem: Inspiring by the following post, I wonder how to solve the recurrence $$ T(n) = 2T(n/2) +\mathcal{O}(n\log n).$$

I had just thought about this question already when I saw the above post. I hope everyone can give me some ideas about this extension version.

Tung Nguyen
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    How is this different from the question you linked? The $O$ on the $n\log n$? It doesn't change the answer, and you should be easily able to fix the proof for this case – Command Master Dec 09 '23 at 03:34
  • @CommandMaster Yes, we can estimate the upper bound of T(n) as $O(n\log{n})$ but how can we estimate the lower bound that deduces the same result $\Theta(n\log^2n)$. – Tung Nguyen Dec 09 '23 at 10:27
  • You can't get a lower bound, because $1=O(n\log n)$, so $T(n) = \Theta(n)$ is also a possible solution – Command Master Dec 09 '23 at 10:38

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