By way of enrichment here is the complexity using Stirling numbers
of the second kind. Using the notation from this MSE
link we have $n$
coupons, and ask about the expected time until a multiset containing
instances of $j$ different coupons has been drawn.
First let us verify that we indeed have a probability distribution
here. We have for the number $T$ of coupons being $m$ draws that
$$P[T=m] = \frac{1}{n^m} \times
{n\choose j-1} \times {m-1\brace j-1} \times (j-1)!
\times (n+1-j).$$
What happens here is that for a run of $m$ samples to produce a
multiset containing instances of $j$ different coupons for the first
time on the last sample we have two parts, a prefix of length $m-1$
and a terminal sample that completes the set. Therefore we must choose
the $j-1$ values excluding the one that occurs last for the prefix
from the $n$ possibilities which gives the first binomial coefficient.
Next we partition the first $m-1$ slots into $j-1$ non-empty sets in
an ordered set partition. (Stirling number and factorial). The
smallest value chosen gets the slots listed in the first set, the next
one those in the second set etc. Finally we get $n-(j-1)$
possibilities ($j-1$ values from the prefix have already been used)
for the terminal sample that completes the selection. Combine with
$n^m$ possible choices.
Recall the OGF of the Stirling numbers of the second kind which says
that
$${n\brace k} = [z^n] \prod_{q=1}^k \frac{z}{1-qz}.$$
This gives for the sum of the probabilities
$$\sum_{m\ge 1} P[T=m]
= {n\choose j-1} (j-1)! (n+1-j)
\frac{1}{n} \sum_{m\ge 1} \frac{1}{n^{m-1}} {m-1\brace j-1}.$$
Focusing on the sum we obtain
$$\sum_{m\ge 1} \frac{1}{n^{m-1}}
[z^{m-1}] \prod_{q=1}^{j-1} \frac{z}{1-qz}
= \prod_{q=1}^{j-1} \frac{1/n}{1-q/n}
\\ = \prod_{q=1}^{j-1} \frac{1}{n-q}
= \frac{(n-j)!}{(n-1)!}.$$
Combining this with the outer factor we get
$${n\choose j-1} (j-1)! (n+1-j)
\frac{1}{n} \frac{(n-j)!}{(n-1)!}
\\ = {n\choose j-1} (j-1)! \frac{(n+1-j)!}{n!} = 1$$
This confirms it being a probability distribution.
We then get for the expectation that
$$\sum_{m\ge 1} m\times P[T=m]
\\ = {n\choose j-1} (j-1)! (n+1-j)
\frac{1}{n} \sum_{m\ge 1} \frac{m}{n^{m-1}} {m-1\brace j-1}.$$
We once more focus on the sum to get
$$\sum_{m\ge 1} \frac{m}{n^{m-1}}
[z^{m-1}] \prod_{q=1}^{j-1} \frac{z}{1-qz}
= \sum_{m\ge 1} \frac{m}{n^{m-1}}
[z^{m}] z \prod_{q=1}^{j-1} \frac{z}{1-qz}
\\ = \left.\left( \prod_{q=0}^{j-1}
\frac{z}{1-qz} \right)'\right|_{z=1/n}
\\ = \left.\left( \prod_{q=0}^{j-1}
\frac{z}{1-qz} \sum_{p=0}^{j-1} \frac{1-pz}{z}
\frac{1}{(1-pz)^2}
\right)\right|_{z=1/n}
\\ = \left.\left( \prod_{q=0}^{j-1}
\frac{z}{1-qz} \sum_{p=0}^{j-1} \frac{1}{z}
\frac{1}{1-pz}
\right)\right|_{z=1/n}
\\ = \prod_{q=0}^{j-1} \frac{1/n}{1-q/n}
\sum_{p=0}^{j-1} \frac{1}{1/n}
\frac{1}{1-p/n}
\\ = \prod_{q=0}^{j-1} \frac{1}{n-q}
\sum_{p=0}^{j-1}
\frac{n^2}{n-p}
= n \prod_{q=1}^{j-1} \frac{1}{n-q}
\sum_{p=0}^{j-1}
\frac{1}{n-p}.$$
Retrieving the outer factor we have
$${n\choose j-1} (j-1)! (n+1-j)
\frac{1}{n} \frac{(n-j)!}{(n-1)!} \times n
\sum_{p=0}^{j-1} \frac{1}{n-p}.$$
The front simplifies to one as before and we are left with
$$n\sum_{p=0}^{j-1} \frac{1}{n-p}
= n \left(\sum_{p=0}^{n-1} \frac{1}{n-p}
- \sum_{p=j}^{n-1} \frac{1}{n-p}\right).$$
This is
$$\bbox[5px,border:2px solid #00A000]{\Large
n \times \left( H_n - H_{n-j} \right)}$$
This yields $n H_n$ when $j = n$ and $1$ when $j=1$ which are both
correct. Using $H_n \sim \log n + \gamma$ we get for $j = n/2$ the
expectation $n\log 2.$
