$[$Using a classic sorting algorithm such as Quicksort would take O$($NlogN$).$ However, in this case, writing a faster sorting algorithm is possible.

Yes, that$$s right. Even though you$$ve learned that the fastest sorts are O$($NlogN$),$ this case is different. Why? In this case, there are limited possibilities for the readings. In such a case, we can sort these values in O$($N$).$ It may be N multiplied by a constant, but that$$s still considered O$($N$).$