Observe that checking every possible interval and every possible grade is too slow for a sufficiently large $N$. Such a solution has a complexity of $\mathcal{O}(N^3)$ and is worth $70\mathrm{\%}$ of points.

To obtain a complete score, we need another approach.

If we assume the grade which the professor will award the students, we can, in a single pass, determine the longest continuous subsequence of desks such that each desk contains at least one student deserving the assumed grade. This can be implemented using a counter storing the current number of continuous desks, and updating it for each desk.

Finally, the solution consists of iterating the described algorithm for each grade from $1$ through $5$ and taking the maximum of the individual results.

Also observe that the problem can be solved with complexity $\mathcal{O}(N)$ even if grades are from the interval $1$ through $N$. Solving this problem is left as an exercise for the reader.