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I'm aware that the worst case scenario recurrence relation corresponding equation is: $$ T(n) = T(n-1) + T(0) + \Theta(n) $$

However, I really don't get how the last term $ \Theta(n)$ was determined.

  1. I dont understand why $ \Theta(n)$ and not $ \mathcal{O}(N) $
  2. I don't get why it is $ \Theta(N) $ and not $ \Theta(N-1) $ because in quick sort you compare the pivot with (n-1) elements, not "n" elements.

Specially the number 2 it's hard for me to understand.

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    $\begingroup$ $\Theta(N)=\Theta(N-1)$ Look at the definition of big-theta. $\endgroup$ Nov 12, 2023 at 15:56
  • $\begingroup$ @RickDecker Oh! I get it, Θ(N) or Θ(N−1) doesn't matter, the constant difference does not affect the growth rate of the function!! Does it makes sense? However, about the question number 1, why Big Thetha and not Big O? $\endgroup$
    – RodrigoAlb
    Nov 12, 2023 at 16:42
  • $\begingroup$ @RodrigoAlb First of all theta, $\Theta(f(n))$ implies $O(f(n))$. Moreover, if the recurrence relation used $O(n)$ instead of $\Theta(n)$ then you would only have been able to conclude that $T(n) = O(n^2)$. This mean that QuickSort, in the worst-case, is not slower than $n^2$ (up to multiplicative and additive constants) but it doesn't rule out, e.g., a time complexity of $O(n \log n)$. By using $\Theta(n)$ you can conlude that $T(n) = \Theta(n^2)$, i.e., there are instances were quicksort uses $\Omega(n^2)$ time. $\endgroup$
    – Steven
    Nov 12, 2023 at 16:52
  • $\begingroup$ Big-O(n) means n or faster. Theta(n) means n. So the worst case is indeed n^2 and not “up to n^2”. $\endgroup$
    – gnasher729
    Dec 13, 2023 at 19:31

1 Answer 1

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  1. For sure you have to do $n-1$ comparisons with the pivot, you can't do less, so it is $\Omega(n-1)=\Omega(n)$. At the same time you do at most $n-1$ comparisons and swaps. Thus it is $O(n-1)=O(n)$. Thus it is $\Theta(n)$.
  2. As stated in 1., summing constants does not affect asymptothic notation, so $\Theta(n)=\Theta(n-1)$.
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