Discussion
The solution of recurrence relation, \(T(n) = 2T(\lfloor \sqrt{n} \rfloor) + \log n\) is
Answer: D
The solution is \(O(\log n \log \log n)\), which is option (D). Put \(m = \log n\); then \(T(2^{m}) = 2T(2^{m/2}) + m\). Writing \(S(m) = T(2^{m})\) gives
\[ S(m) = 2S(m/2) + m \]
which, by the master theorem, is \(\Theta(m \log m)\). Going back to \(n\):
\[ T(n) = \Theta(\log n \log \log n) \]
The bounds with a factor n in (A) and (B) are far too large, and \(O(\log \log n)\) in (C) is too small, since the term \(\log n\) is added at the top level.
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