That statement is only very superficially similar.

Counting them is saying how many of them there are, it doesn't necessarily
require you to look at/compute each one.
So it is not the same as printing them.

If you're saying I want to print out each inversion individually then it's
going to be n^2 and there's no way around it.
If you have some other encoding so you can print something which represents
multiple inversions in 1 step, you might be able to get under n^2.


On 22 October 2012 12:03, Aamir Khan <syst3m.w...@gmail.com> wrote:

>
>
> On Monday, October 22, 2012, Dipit Grover wrote:
>
>> Since the number of inversions are of order n^2 in the worst case, so
>> should be the complexity of printing them apparently.
>
>
> It makes sense to some extent but this is no proof. There has to be a
> better proof for lower bound of complexity for this algorithm. Because I
> can state similar statement,
>
> "Since the number of inversions are of order N^2 in the worst case, so
> should be the complexity of counting them apparently".
>
> But we all know this is not true as we already know O(nlogn) solution.
>
>
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> Aamir Khan | 4th Year  | Computer Science & Engineering | IIT Roorkee
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