Analysis

To analyze the amortized cost of the ith TableInsert operation, let

$\;\;\;\;\;$num_i denote the number of items stored in the table after the ith operation

$\;\;\;\;\;$size_i denote the total size of the table after the ith operation

$\;\;\;\;\;$ denote the potential after the ith operation

Initially, num₀ = 0, size₀ = 0, and = 0.

Consider two cases based on whether a table expansion is done.

No expansion:

= 1 + (2*num_i-size_i) - (2*num_i-1 - size_i-1)

= 1 + (2*num_i-size_i) - (2*(num_i - 1) - size_i)

= 1 - (-2) = 3