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.
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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.