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When using Dijkstra's algorithm with a priority queue, the time complexity is O((V + E) log V) because of the following reasons:
Initialization: In the beginning, each vertex in the graph is added to the priority queue, which has a cost of O(V) due to adding all vertices to the queue.
Main loop: The main loop of Dijkstra's algorithm runs for V iterations, where V is the number of vertices in the graph. In each iteration, we remove a vertex from the priority queue based on its minimum distance value, which takes O(log V) time due to the use of a priority queue.
Relaxation: For each edge in the graph (E total edges), we check if updating the distance value of its neighbor vertex would result in a shorter path. If so, we update the distance value and reposition the vertex in the priority queue based on its new distance value. This operation also takes O(log V) time.
Overall complexity: Combining all these steps, we get a total time complexity of O((V + E) log V) for Dijkstra's algorithm with a priority queue.
Therefore, with these considerations and operations within Dijkstra's algorithm with a priority queue implementation result in this specific time complexity.
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