Dijkstra’s shortest path algorithm is O(ElogV) where:
V is the number of vertices
E is the total number of edges
Your analysis is correct, but your symbols have different meanings! You say the algorithm is O(VElogV) where:
V is the number of vertices
E is the maximum number of edges attached to a single node.
Let’s rename your E to N. So one analysis says O(ElogV) and another says O(VNlogV). Both are correct and in fact E = O(VN). The difference is that ElogV is a tighter estimation.
Adding a more detailed explanation as I understood it just in case:
O(for each vertex using min heap: for each edge linearly: push vertices to min heap that edge points to)
V = number of vertices
O(V * (pop vertex from min heap + find unvisited vertices in edges * push them to min heap))
E = number of edges on each vertex
O(V * (pop vertex from min heap + E * push unvisited vertices to min heap)). Note, that we can push the same node multiple times here before we get to “visit” it.
O(V * (log(heap size) + E * log(heap size)))
O(V * ((E + 1) * log(heap size)))
O(V * (E * log(heap size)))
E = V because each vertex can reference all other vertices
O(V * (V * log(heap size)))
O(V^2 * log(heap size))
heap size is V^2 because we push to it every time we want to update a distance and can have up to V comparisons for each vertex. E.g. for the last vertex, 1st vertex has distance 10, 2nd has 9, 3rd has 8, etc, so, we push each time to update
O(V^2 * log(V^2))
O(V^2 * 2 * log(V))
O(V^2 * log(V))
V^2 is also a total number of edges, so if we let E = V^2 (as in the official naming), we will get the O(E * log(V))