instead of the n lg n that using a hash provides
You are wrong here. The expected amortized time for hash operations is O(k), where k is the length of the string (it's the time that's needed to compute the hash value). Note the absense of a dependency on the size of a hash. This means that if you do N hash operations (insertions, deletions, fetches) and all your strings are bounded in length by some constant, your expected time is O(N).

Now, an individual insertion in a hash may take O(n) (where n is the size of the hash). But this typically happens only after Ω(n) insertions (due to the hash filling up and a new hash needs to be build), so that amortizes out.

An individual search may take Ω(n) time because a fraction of your keys hash to the same bucket. That would lead to a worse case quadratic performance. But the chances of that happening are so small, it doesn't have an impact on the expected running times. And ever since 5.8.1, Perl has some safety mechanisms (turned on by default) that even prevents you from constructing an input sequence that hashes the keys to the same bucket, because that hash function uses a different values on each run.

So, for all practical purposes, hash operations are done in constant time.

In reply to Re^2: better union of sets algorithm? by Anonymous Monk
in thread better union of sets algorithm? by perrin

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