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\%~ 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.