My first instinct was to just simulate it naively, count all substrings after each replacement, and check against m.
Model the problem as counting substrings that contain at least one '*' after each attack, and find the earliest time when this count reaches m. Use a data structure like a sorted set of attacked positions to efficiently compute the number of substrings without '*' and subtract from total substrings. Alternatively, binary search on time t and check if the condition holds, using a linear scan to count substrings with '*'.
Pro tip: Clarify edge cases upfront: if m is 0 or the initial string already has enough substrings with '*' (impossible since no '*' initially), handle accordingly. Also, consider that the answer might be n+1 if never reached, but the problem likely guarantees it will be reached.
Restate the problem: after each attack, some positions become '*'. We need the number of substrings that contain at least one '*'. This equals total substrings minus substrings consisting only of non-'*' characters. Total substrings = n*(n+1)/2.
Since n can be large, we need an efficient way to update the count after each attack. Use a sorted set (e.g., balanced BST) to maintain the positions of '*'. When a new position is attacked, it splits an existing segment of non-'*' characters into two, reducing the count of substrings without '*' by the sum of substrings in the old segment minus the sum in the two new segments.
Initialize the sorted set with sentinel positions -1 and n to represent boundaries. Initially, there are no '*', so the number of substrings with '*' is 0. For each attack at position p (in the given permutation order), find the predecessor and successor in the set. The segment between them (exclusive) was all non-'*'. After inserting p, this segment splits into two. Update the count of substrings without '*' accordingly, then compute substrings with '*' = total - without. Check if it reaches m.
If the initial count (0) already >= m, return 1 (but m is likely >0). Otherwise, after each attack, if the count >= m, return the current time t (1-indexed). If after all attacks it never reaches m, return -1 or as specified (but problem implies it will).
Time complexity: O(n log n) due to n insertions and predecessor/successor queries in a balanced BST. Space: O(n) for the set. Mention that a binary search approach with O(n) check per step would be O(n log n) as well but might be simpler to implement; however, the incremental approach is more direct.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.