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I'm having difficulties with analyzing the worst-case runtime of this following case:

I'm given an array that has $n$ natural numbers. Out of all $\binom{n}{2} = \frac{n(n-1)}{2}$ possible pairs there are $\frac{n(n-1)}{2} - 5n$ inversions. Inversion is defined like this: $a[i] < a[j],\ i>j$.

So of course you can sort the array in $O(n \log n)$ time but can you do any better than this? I've looked online and saw that inversion sort time is in $\Theta(n+I)$ where $I= \text{# inversions}$ but then I get that $I=\Theta(n^2)$ so it doesn't really help me.

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Hint: Make use of the following facts:

  • Each pair $i,j$ that forms an inversion for sorting in ascending order, does not form an inversion when sorting in descending order (and vice versa).
  • Reversing the complete list can be done in $\Theta(n)$.
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  • $\begingroup$ thank you very much I actually figured that out a few minutes ago. $\endgroup$ – Jake Keen Jul 10 '14 at 9:55

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