In order to score ~20\%~, it was enough to simply simulate the process described in the problem statement.

The time complexity of that solution is ~\mathcal O(NQN) = \mathcal O(QN^2)~.

In order to score an additional ~20\%~, you could use prefix sums to speed up the solution above. Suppose that for every letter you built an array ~pref~ in which an element at position ~i~ stores the number of occurrences of that letter from the beginning of the string to the ~i^\text{th}~ letter (inclusive). For example, for a word abcaab and a letter a, the array ~pref~ would be ~[1, 1, 1, 2, 3, 3]~. That array is useful because now we can determine in ~\mathcal O(1)~ what is the number of occurrences of a certain letter in an interval from ~L^\text{th}~ to the ~R^\text{th}~ letter in our original word. The formula for that is ~pref[R]-pref[L-1]~.

The idea is now to precompute this array for every letter, thus obtaining an algorithm with complexity ~\mathcal O(N(N\Sigma \times Q\Sigma)) = \mathcal O((N^2+Q)\Sigma)~, where ~\Sigma~ denotes the size of our alphabet.

In order to score all points, you could observe the following: if a word is perfect after ~i^\text{th}~ snowball, it will also be perfect after every other snowball thrown after the ~i^\text{th}~ one. This property allows us to use binary search. The check inside binary search is done similarly to the algorithm above (using prefix sums).

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