Q:
Average-case?
A:
Let
n
=
Number of keys stored in
T
m
=
Number of slots in
T
α
=
n/m
Note:
α
= average number of elements in a chain, or
load factor
α
can be less than or greater than
1
If
m
is proportional to
n
(that is,
m
is chosen as a linear function of
n
), then
n = O(m)
.
In that case,
α = n/m = O(m)/m = O(1)