DFS runs in O(n + m) time provided the graph is represented by the adjacency list structure; Recall that Σv deg(v) = 2m; BFS(analysis): Setting/getting a vertex/edge label takes O(1) time; Each vertex is labeled twice once as UNEXPLORED; once as VISITED; Each edge is labeled twice once as UNEXPLORED; once as DISCOVERY or CROSS

Some confusion about time-complexity and A*. According to A* Wiki the time-complexity is exponential in the depth of the solution (shortest path): The time complexity of A* depends on the heuri...

3. When graph go denser ( Worst case is Complete Graph ) we use Fibonacci Heap and adjacency list: O( e + v log v) Time complexity of kruskal is O(e log e) in Worst case e ~ v^2 so log (v^2) = 2 log v. So we can safely say than O(e log e) …

For instance if you store the adjacency list as a map of lists the time complexity is O(E) for exactly the reasons you mention. It is the best time complexity you can get for this. But if you use a list of lists you might end up implementing a O(EV) time complexity (e.g.: going through V vertices to check if the tail vertex exists for each edge ...

Nov 4, 2017 · What is the time complexity of search query in Graph database (especially Neo4j) ? I'm having relational data with me. I'm confused to use a Relational database or Graph database to store that data. So, I want to store the data based on the performance and time complexity of the queries for that particular database.

Nov 6, 2014 · This time complexity of this find operation depends on the data structure you use; if you use a HashMap, it will be O(1); if you use a List, it will be O(V). Once you have identified the vertex that needs to be removed, you now need to remove all the edges of that vertex.

Nov 19, 2015 · I don't mean graphically similar (to a straight line) but time-complexity similar. O(nlogn) time can easily be an order of magnitude bigger than O(n). If the graphs compared O(nlogn) and O(n) algorithms you would see what I mean.

Apr 6, 2020 · Because of that we can reduce the time complexity estimate O(V+E) of the cycle detection algorithm to O(V). Starting the DFS from all vertices of the graph is necessary in the case when the graph consists of a number of connected components - the "visited" boolean variable guarantees that the DFS won't traverse the same component again and again.

Apr 7, 2016 · Time Complexity: If you can access each node in O(1) time, then with branching factor of b and max depth of m, the total number of nodes in this tree would be worst case = 1 + b + b 2 + … + b m-1. Using the formula for summing a geometric sequence (or even solving it ourselves) tells that this sums to = (b m - 1)/(b - 1), resulting in total ...

Jun 27, 2016 · I want to point out that this time complexity, O(E log V), assumes the given graph is connected. In the case of a sparse graph that has a lot of lone vertices, for example, it will not hold. That is why the worst case for Dijkstra binary heap implementation is O(V log V + E log V).