Subtask 1

We can directly simulate excluding each string individually. For each string, loop through all the other strings and check if all characters appear at least once.

Each of the ~S~ strings we consider excluding, and we must loop through the remaining ~S-1~ strings with length ~L~ yielding a time complexity of ~\mathcal{O}(S^2 \times L)~.

Subtask 2

As ~L~ is now larger, we need to find a better way than looping all ~L~ characters when looking through a string.

Since we only care about whether or not a string contains a character, rather than the number of them, we can simply create a boolean array table for each character for each string. If we loop through each string, and each string mark down in the boolean table whether or not a character has been seen, later when looping through a string we can simply refer to the boolean table of only ~26~ elements rather than looping through the string of length ~L~. With this preprocessing, this yields a time complexity of ~\mathcal{O}(S^2 + S \times L)~.

Full Solution

As ~S~ is now larger, we need a faster way than simulating excluding each string.

Before excluding any string, first check if all the strings together are a pangram. If any character doesn't appear in any of the strings, we know there are ~0~ ways to create a pangram with ~1~ less string.

Otherwise, we can observe that if a string were to be excluded and the remaining strings do not form a pangram, then there must be at least one character that exclusively shows up in that string. We can build a frequency table of size ~26~, indicating the number of strings that contain each character. Afterwards, we can loop through each string, and if we find a string that has a character that only appears in one string, then we know we must keep that string.

We can assume all ~S~ strings could be excluded, then subtract off any ones we know must be kept, and that leaves us with our final answer. This yields a time complexity of ~\mathcal{O}(S \times L)~.

Alternative Solution

Alternatively, we can speed up the subtask ~2~ solution by ordering the strings and maintaining a prefix and suffix frequency table of all characters seen. If we check if the ~i^\text{th}~ string can be excluded, the combined frequency tables of the first ~i-1~ strings and all strings after ~i~ shouldn't have any characters with no frequency. This yields the same time complexity as of ~\mathcal{O}(S \times L)~.