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OmG
OmG
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As n^(0.5)$n^{0.5}$ is always greater than log(n)$\log(n)$, O(n^(2.5))=O((n^2)*(n^0.5))$O(n^{2.5})= O(n^2 \times n^{0.5})$ is always bigger than O((n^2)*log(n))$O(n^2 \times \log(n))$. Anyway, you should consider your real algorithm usage scenario to choose one which fits the best.

As n^(0.5) is always greater than log(n), O(n^(2.5))=O((n^2)*(n^0.5)) is always bigger than O((n^2)*log(n)). Anyway, you should consider your real algorithm usage scenario to choose one which fits the best.

As $n^{0.5}$ is always greater than $\log(n)$, $O(n^{2.5})= O(n^2 \times n^{0.5})$ is always bigger than $O(n^2 \times \log(n))$. Anyway, you should consider your real algorithm usage scenario to choose one which fits the best.

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Farhad Rahmanifard
Farhad Rahmanifard

As n^(0.5) is always greater than log(n), O(n^(2.5))=O((n^2)*(n^0.5)) is always bigger than O((n^2)*log(n)). Anyway, you should consider your real algorithm usage scenario to choose one which fits the best.