Worst Case Scenario for Quicksort algorithm with pivot element n/2 What would the worst case array look like if I decide to always take the element on the position $\frac{n}{2}$ as the pivot element?
I know that if I choose the left or rightmost element as pivot ,the worst case occurs if:


Array is already sorted in same order
Array is already sorted in reverse order
All elements are same


and that the complexity in that cases is $\mathcal{O}({n^2})$.
However, this cases should not be a problem if I take the middle index of the partition as my pivot element.
Since you're not guaranteed anything about the order of the input, it's possible that the n/2 position has the smallest/largest element of the input. Then quicksort will proceed and put everything else on one side of the pivot.