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If you have this expression:

$$\frac{1+n+\left\lceil\log_2{n}\right\rceil}{1+\left\lceil\log_2{n}\right\rceil}$$

How do you obtain a big $O$ from this? I think it's just $O(\log_2n)$. Is this right?

Adi Dani
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omega
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1 Answers1

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$$\frac{1+n+\left\lceil\log_2{n}\right\rceil}{1+\left\lceil\log_2{n}\right\rceil} = \frac{O(1+n+\left\lceil\log_2{n}\right\rceil)}{\Omega(1+\left\lceil\log_2{n}\right\rceil)} = \frac{O(n)}{\Omega(\log n)} = O\left(\frac{n}{\log n}\right)$$

dtldarek
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