I believe you mean "the number of buckets" not the number of valid hash keys ... there is a catch though ... the m usually depends on n! How exactly, depends, but I beleive in the Perl builtin implementation its 2*n rounded to the nearest power of 2. Which means your o(n/2m) becomes o(n/4n) wich becomes o(1/4) and ... well ... constants are irrelevant in the o() notation thus it's o(1). QED.
The average case of hash lookup even with the naive implementation is o(1), the worst case is O(n) for the naive implementation, but there are ways around that. They make the implementation more complex and insertion more costly, but they are possible. You can rehash once the list in a bucket gets too long or you can use binary trees instead of lists in the buckets.
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