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copied!<p>Well, not sure if it's better but I had fun writing this tree. </p> <p>It uses dictionnary for nodes (should use something better as I use the tag "data" which is unclean)</p> <p>Root is root of tree, it gets filled by parsing initial data</p> <p>About complexity, it's unclear what is n : </p> <ul> <li>if you have lots of project (x) with few tags per project (y) then you want a good complexity on x and you don't care about y</li> <li>if you have a few projects with lots of tags, then you want the contrary</li> </ul> <p>I suppose first solution is mote common.</p> <p>I I'm not mistaken, my solution runs in x.y.ln(y) (I need to sort tags per project) to create the tree.</p> <p>Displaying the tree is y.y (but uses recursion and a temporary list so it must bloat memory).</p> <pre><code>d= { "apple": ["fruit", "green", "nature"], "sometree": ["tree", "green", "wood", "nature"], "banana": ["fruit", "nature"], "someplant": ["green", "wood", "nature"], "otherplant": ["green", "wood", "nature"] } root=dict() root['data']=[] for k,v in d.iteritems(): v.sort() # r=root for c in v: try: r=r[c] except KeyError: r[c]=dict() r=r[c] r['data']=[] r['data']+=[k] def dump(r,hist): result=r['data'][:] for x in r.keys(): if x=='data': continue result+=dump(r[x],hist[:]+[x]) if len(result)>0 and len(r['data'])>0: print (hist,result) return result dump(root,[]) </code></pre> <p>Code is far from perfect (quick and dirty mode) but it looks like it works:</p> <pre><code>$ c:\Python27\python.exe temp.py (['green', 'nature', 'tree', 'wood'], ['sometree']) (['green', 'nature', 'wood'], ['otherplant', 'someplant']) (['fruit', 'green', 'nature'], ['apple']) (['fruit', 'nature'], ['banana']) </code></pre>

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