# 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:

1. Array is already sorted in same order
2. Array is already sorted in reverse order
3. 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.