re: Daily Challenge #3 - Vowel Counter VIEW POST

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re: I don't think so. Consider pseudocode like this: if (letter == 'a' || letter == 'e' || letter == 'i' || letter == 'o' || letter ...

You also have to traverse the tree to get to the next letter. But I see what you mean. Only counting equality checks, hard-coding behaves like a linear search. My bad.

I think you might still be missing Brian's point that a hash set has O(1) lookup time, which is faster than the tree set's O(log n).

(On paper. Real-world implementations vary. e.g. Ruby's hashes just do linear search at this size.)

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