Analysis

Thus, for all buckets,

$$\sum_{i=0}^{n-1} O(E(n_i^2)) \;=\; O(\sum_{i=0}^{n-1} E(n_i^2)).$$

Given:

• n elements, uniformly distributed over all possible values
• n buckets

What is the probability that an element will be inserted into some bucket i?

Answer: p = 1/n.

Given n trials consisting of putting elements into buckets, how many elements will be inserted into bucket i?

n_i = Binomial(n,p)

with mean E(n_i) = np = 1

and variance Var(n_i) = np(1-p) = 1 - 1/n

E(n_i²) = 1 - 1/n + 1²

= 2 - 1/n

= $\Theta(1)$

Therefore, T_IS = $\Theta(1)$
and T_BS = 4n + $\Theta(1)$n + 3 = O(n) for the average case. In the worst case, the run time is n².