I'm not sure I understand the answer to this question:
What is the running time of Dijkstra's algorithm in a graph that is sufficently sparse - in particular, $E=o(V^2/\log V)$, where $V$ is the number of vertices and $E$ the number of edges. Assume that we are implementing the min-priority queue with a binary min-heap. Choose the anwser the that gives the tightest correct bound.
- $O((V+E)\log V)$
- $O(V\log V)$
Answer: 4 $O((V+E)\log V)$
As I understand it, the algorithm runs in $O((E+V) \log(V))$ with binary heap. Thus when, $E = V^2/\log(V)$, shouldn't the overall time complexity be $O(V^2)$?