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nir shahar
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Neither of those. For $n>1$, $\lfloor\frac{1}{n}\rfloor=0$. Hence, $\lfloor\frac{1}{n}\rfloor=0=O(0)$.

Also, you can easily rule out $\Omega(\log(n))$ since $\log$ is an increasing function, while $\frac{1}{n}$ is a decreasing function.

nir shahar
  • 11.7k
  • 3
  • 16
  • 35