A solution that, for each string, iterates over the previous ~K~ strings and counts the strings with the same number of letters, is too slow.

A faster solution reads a string, counts its letters – let us denote the number of letters with ~L~ – and then immediately, without another for loop, answers the question: how many strings, out of the previous ~K~, have exactly ~L~ letters?

In order to find that number quickly, we need to keep an auxiliary array with the corresponding count for each ~L~. This array must, of course, be maintained: upon reading a new string with length ~L~, we increment the ~L^\text{th}~ element of the auxiliary array by one, and decrement the ~(L')^\text{th}~ element by one, where ~L'~ is the length of the string "falling out" of the last ~K~ strings interval, i.e. not included in the set of friends of upcoming strings anymore.

This solution is actually based on the sweep-line principle: the imaginary scanner is scanning through the rankings list and processing events such as new name added and name removed from friends set.