For the first subtask, it suffices to maintain an array ~a~, where ~a[i]~ represents the opinion of the ~i^\text{th}~ classmate. To handle updates, simply increment/decrement the given index, and to handle range queries, loop over the range and count how many of the specified values exist in that range.

Time complexity: ~\mathcal O(NQ)~

For full marks, we maintain a segment tree where each node contains a map which maps the opinions in that range to their count. Because all values are initialized to ~0~, and we only have to handle increment and decrement updates, there can be at most ~1413~ distinct values in the whole segment tree at any given time, which is enough to justify the overhead of having a map at each node. However, to leverage this, special attention must be given to delete keys in the map which are not in use.

Time complexity: ~\mathcal O((N+Q) \log N)~