The recursion is:

T(n) = 5T(n/2) + O(n)

I solved for the time complexity using Master theorem and found Θ(n^2). but, the question has asked to find the upper bound in terms of Big Oh. Is there any principle to convert it into Big Oh form? I read some examples where they substituted some values and deduced but I couldn't understand.

Update: I apologize for writing the recursion equation wrong. The actual recursive equation for this question is:

T(n) = 2(n-1) + O(n)
  • $\begingroup$ $\Theta$ is subset of $O$, so if you obtain first, then you have, also, second. $\endgroup$
    – zkutch
    Sep 26 '21 at 21:11

Your reasoning is wrong. It is in $\Theta(n^{\log_2(5)})$. Hence, it is also in $O(n^{\log_2(5)})$.

Answer to the update: Also, for the update part, it is wrong. You can find it by a straightforward expansion (no need to master theorem):

$$ T(n) = 2T(n-1) + O(n) = 2(2T(n-2) + O(n-1)) + O(n) \\ = 2^2T(n-2) + O(n-1) + O(n) = \cdots = \\ 2^n * T(0) + O(1) + O(2) + \cdots + O(n) \in O(2^n) $$

  • $\begingroup$ I just realized that I typed the wrong expression. Thanks for the answer. $\endgroup$ Sep 27 '21 at 0:34
  • $\begingroup$ I will accept your answer once you make corrections. Sorry for that. $\endgroup$ Sep 27 '21 at 0:49
  • $\begingroup$ @five_star_021 You shouldn't totally change the question to another question. You should post a new question for that matter. $\endgroup$
    – OmG
    Sep 27 '21 at 11:12
  • $\begingroup$ @five_star_021 it's better now. You can find the answer for the updated part as well. $\endgroup$
    – OmG
    Oct 3 '21 at 14:31

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