Interval Diversity

Fix the left endpoint $l$, and extend the right endpoint $r$ from $l$ to $N$ one by one.

Maintain the occurrence counts of values inside the current interval. When $A_r$ is added, we can check whether this value was absent before. If so, the number of distinct integers $x$ increases by $1$.

The interval length is $y=r-l+1$, so we compute $x^2/y$ for every interval and update the maximum.

There are $O(N^2)$ intervals, and each update can be done in expected $O(1)$ time. Therefore, the total time complexity is $O(N^2)$.