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    <p>This problem can be reduced to a problem in graph theory where you find all groups of connected nodes in a graph.</p> <p>An efficient way to solve this problem is doing a "flood fill" algorithm, which is essentially a recursive breath first search. This <a href="http://en.wikipedia.org/wiki/Flood_fill" rel="nofollow">wikipedia entry</a> describes the flood fill algorithm and how it applies to solving the problem of finding connected regions of a graph. </p> <p>To see how the original question can be made into a question on graphs: make each entry (e.g. "Server1", "Server_1", etc.) a node on a graph. Connect nodes with edges if and only if they are synonyms. A matrix data structure is particularly appropriate for keeping track of the edges, provided you have enough memory. Otherwise a sparse data structure like a map will work, especially since the number of synonyms will likely be limited. </p> <ul> <li>Server1 is Node #0 </li> <li>Server_1 is Node #1</li> <li>Server_2 is Node #2</li> </ul> <p>Then edge[0][1] = edge[1][0] = 1, indicated that there is an edge between nodes #0 and #1 ( which means that they are synonyms ). While edge[0][2] = edge[2][0] = 0, indicating that Server1 and Server_2 are <strong>not</strong> synonyms.</p> <p><strong>Complexity Analysis</strong></p> <p>Creating this data structure is pretty efficient because a single linear pass with a lookup of the mapping of strings to node numbers is enough to crate it. If you store the mapping of strings to node numbers in a dictionary then this would be a O(n log n) step. </p> <p>Doing the flood fill is O(n), you only visit each node in the graph once. So, the algorithm in all is O(n log n).</p>
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    1. COYes, that seems to be an option. Thank you!
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    2. COIf algorithm is O(n) and you have to visit every node once, then overall complexity is O(n^2). If you tries with graphs, then some closure calculation where more appropriate.
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    3. CO@p4553d: I don't follow why it is O(n^2). You visit each node once and you are done - as soon as you have visited a node you know which sub-graph it belongs to. The key is of course to skip over nodes that you have already visited, which the flood fill algorithm already does.
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