Discussion
A hash table with 128 buckets holds 512 entries using separate chaining. What is the average chain length?
Answer: D
Correct answer: 4
\[ \frac{512}{128} = 4. \] This is the load factor, and it is what governs lookup cost under separate chaining: each search examines about four entries on average. Doubling the number of buckets would halve it, which is why hash tables are grown as they fill.
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