In the ordinary coupon collector's problem, there are $n$ types of coupons, and we pick coupons uniformly at random until we get at least one of each coupon.
https://en.wikipedia.org/wiki/Coupon_collector's_problem
It is well known that the expected time to collect all coupons $k$ times (for a fixed integer $k$) is $ n \log n + (k - 1) n \log\log n + O(n)$. But what if we need to collect all coupons $n$ times? Is the expected time $\omega(n^2)$?
Following Balls into Bins: A tight analysis assume we throw $m$ balls independently and uniformly into $n$ bins and let $M$ be the maximum number of balls in any bin. This allows for $m$ to depend on $n$. Then, if $m\gg n (\log n)^3,$ the probability $P[M>k_\alpha]=o(1)$ if $\alpha>1$ and $P[M>k_\alpha]=1-o(1)$ if $\alpha\in (0,1).$ Here $$ k_\alpha=\frac{m}{n}+ \sqrt{\frac{2 m \log n}{n}\left(1-\frac{1}{\alpha}\frac{\log \log n}{2 \log n}\right)}. $$ This means that there is threshold behaviour for the maximum load around $m/n,$ and it will be upper bounded by $k_\alpha$ with probability going to one if we take $\alpha>1.$ Now let $m=n^2,$ and note that the bounded maximum implies that at least $n^2$ time is needed for the minimum load to exceed $n.$ In other words, since the maximum load is very close to the average load so is the minimum load.