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s_tim

How does this compare to Enzyme (https://enzyme.mit.edu/)?
But you would need to spend the $15 on every request whereas the RAG approach would be most likely significantly cheaper per request.
Re:The Programming Interview from Hell on Monday February 20, 2017 @10:23AM (#13687980)
You build from the different tuples in your list just one graph. Then it's just a simple DFS/BFS with one random start node. Which gives you your first component. Then you can get the second if you start at a node which is not in the previous component until you visited all nodes. This should all be in O(n).
Re:The Programming Interview from Hell on Monday February 20, 2017 @09:52AM (#13687692)
You could model the problem as a graph (each integer represents a vertex and two consecutive integers an edge, e.g. (1, 2, 3) is a graph with nodes 1,2,3 and edges between 1 and 2 and 2 and 3). Then your problem is just to find all connected components of the graph (https://en.wikipedia.org/wiki/Connected_component_(graph_the....