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by kragen 5422 days ago
I suppose that depends on what you store in your trie. If every node contains a field for the length of the longest path below it, then yes. But that's not a normal thing to store in a trie, and it wasn't at all obvious to me that that was what you meant.
1 comments

Not true.

If you keep on following the trie, you'll fall out of the trie at the point that there are no words that have that sequence at the start. Which will happen by the time you exceed the longest word in the dictionary, but will probably happen substantially before that.

Am curious to see having the dictionary in a trie allows you to build an implementation that is O(n). Will add it to the blog post if it works.