For a given graph G = (V,E) how can you sort its adjacency list representation in O(E+V) time?

Question

Because we know that the integers representing a vertex can take values in [0,...,|V|-1] range, we can use counting sort in order to sort each entry of the adjacency list in O(V) time.

Since we have V lists to sort, that would give us a O(V^2) time algorithm. I don't see how we can transform this into an O(V+E) time algorithm...

Answer 1

In fact you need to sort E elements in total - the number of edges. Thus your estimation of O(V^2) is not quite correct. You sort each of the adjacency lists in linear time with respect to the number of edges it contains . And as in total you will have E edges, the complexity of sorting all lists will be O(E) . Of course as you have V lists, you can't get lower than O(V) and thus the estimation O(V +E) .