计数排序的下限为 O(n)

Question

The running time of counting sort is Θ (n+k). If k=O(n), the algorithm is O(n). The k represents the range of the input elements. Can I say that the Counting sort has a lower bound of O(n) because the algorithm takes O(n) time to compute a problem and that the lower bound of O(n) shows that there is no hope of solving a specific computation problem in time better than Ω(n)??

Answer 1

Well yes since T(n,k) = Theta(n+k) then T(n,k) = Omega(n+k). Since k is nonnegative we know that n + k = Omega(n) and so T(n, k) = Omega(n) as required.

Answer 2

Another perspective on why the lower bound is indeed Ω(n): if you want to sort an array of n elements, you need to at least look at all the array elements. If you don't, you can't form a sorted list of all the elements of the array because you won't know what those array elements are. :-)

That gives an immediate Ω(n) lower bound for sorting any sequence of n elements, unless you can read multiple elements of the sequence at once (say, using parallelism or if the array elements are so small that you can read several with a single machine instruction.)