![]() ![]() Space complexity of Dijkstra’s algorithm is O(V2) where V denotes the number of vertices (or nodes) in the graph. To improve performance, priority queues are typically based on a heap, giving O(log n) performance for inserts and removals, and O(n) to build the heap. Time complexity of Dijkstra’s Algorithm Space complexity of Dijkstra’s algorithm When using binary heaps, the average case time complexity is lower than the worst-case: assuming edge costs are drawn independently from a common probability distribution, the expected number of decrease-key operations is bounded by O(|V| * log(|E| / |V|)), giving a total running time of O(|E| + |V| * log |E| / |V| * log |V|).The Fibonacci heap improves this to O(|E| + |V| * log |V|). The worst case (where log denotes the binary logarithm log 2 ) for connected graphs this time bound can be simplified to Θ (|E|log |V|).Best and Worst Cases for Dijkstra’s Algorithm The reason is, Fibonacci Heap takes O(1) time for decrease-key operation while Binary Heap takes O(Logn) time. ![]() Time complexity can be reduced to O(E + VLogV) using Fibonacci Heap. So overall time complexity is O(E+V)*O(LogV) which is O((E+V)*LogV) = O(ELogV) Note that the above code uses Binary Heap for Priority Queue implementation. The inner loop has decreaseKey() operation which takes O(LogV) time. If we take a closer look, we can observe that the statements in inner loop are executed O(V+E) times (similar to BFS). The time complexity of the given code/algorithm looks O(V^2) as there are two nested while loops.
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