Subtask 1

For every query, iterate through all the colours and find the first occurrence of ~a_i~ and the last occurrence of ~b_i~.

The answer is given by ~\max_{1 \le i \le M} \{\mathrm{last}(b_i) - \mathrm{first}(a_i) \mid b_i\text{ and }a_i\text{ are colours that exist in the scarf}\}~.

Time complexity: ~\mathcal O(NM)~

Subtask 2

There is no intended solution. (Red herring task.)

Time complexity: ~\mathcal O(NM)~

Subtask 3

Keep track of the first and last occurrence of every colour. For every query, it's now possible to query ~\mathrm{first}(i)~ and ~\mathrm{last}(i)~ in ~\mathcal O(1)~. The rest of the solution follows from Subtask 1.

Time complexity: ~\mathcal O(N+M)~