Correct : a

While choosing the central element often prevents the O(n²) case on already sorted data, a specific input sequence can still be constructed (an adversary array) that forces the central element to consistently be the smallest or largest in every sub-array.

This results in the most unbalanced partitions: one sub-array of size n-1 and one of size 0.

The recurrence relation for this worst case is:

T(n) = T(n-1) + O(n)

This solves to O(n²).