Analysis of separatechaining hashing

Keep in mind the
load factor
measure of how full the table is:
α
= N/M
where M is the size of the table, and
N is the number of keys that have been inserted in the table

With separate chaining, it is possible to have
α > 1

Given a load factor
α
, we would like to know the time costs, in the best, average, and worst case of

newkey insert and unsuccessful find (these are the same)

successful find

The best case is O(1) and worst case is O(N) for all of these... let’s analyze the average case
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