In worst case, we may pick pivot elements in the increasing order (input also given in sorted order) which will result in running time of O($n^{2}$)
Both the deterministic and randomized quicksort algorithms have the same best-case running times of O($nlogn$) and the same worst-case running times of O(n$^{2}$).The difference is that with the deterministic algorithm, a particular input can elicit that worst-case behavior. With the randomized algorithm, however, no input can always elicit the worst-case behavior. The reason it matters is that, depending on how partitioning is implemented, an input that is already sorted--or almost sorted--can elicit the worst-case behavior in deterministic quicksort.
source: Thomas Coremen
Ans. C