Suppose we are given n keys, m hash table slots, and two simple uniform hash functions h1 and h2. Further suppose our hashing scheme uses h1 for the odd keys and h2 for the even keys. What is the expected number of keys in a slot?

A.

mn

B.

nm

C.

2nm

D.

n2m

Solution:

Load Factor (α) = Number of keys / Number of slots in the hash table.

So, the number of expected keys in a slot is nm.

Option B is correct.