When average , worst and best case time complexity happens in quick sort? I know recurrence relation corresponding to quick sort worst case is 

$T(n)=T(n-1)+T(0)+\Theta(n)$

and time complexity is $O(n^2)$.

This happens when we select pivot which is either largest element or smallest element in the current sub problem.

This can occur when array is sorted in either descending or ascending order and we select either largest or smallest element as pivot.

I also know that best case occurs when median is chosen as pivot, giving recurrence relation

$T(n)=T(n/2)+T(n/2)+\Theta(n)$.

and time complexity $O(n \log n)$.


  Statement 1: Any other partition should result in average case. 


I have come across problems each time choosing pivot which partition array in some fixed proportion (say $T(n)=T(n/10)+T(9n/10)+\Theta(n)$) ending in average case performance of $O(n \log n)$ (and I understand its proof). 

However I have also came across text saying:


  Statement 2: Any pivot which does not partition array in some proportion should result in worst case. If $n$th smallest or largest element is selected as pivot, it will result in worst case. 


Q1. Is statement 1 true? If no, which other pivot selection strategies could result in worst case or best case behavior?

Q2. Is statement 2 true?
Remember, for big-O we care about the asymptotic limit. If you choose the $k$th largest (or smallest) element, for constant $k$, then when you make $n$ big enough it's as though you chose the single largest or smallest. The $k$ elements on the other side just aren't enough to matter any more. So statement 2 is true.

Statement 1 is also true; you say you understand the proof of it, so I'm not sure how much detail to go into here. But intuitively, the amount on each side of the pivot grows proportionally to $n$. That's what makes it different from the constant $k$.