This is a binary search on the answer combined with an efficient method to count distinct substrings containing at least one '*'. For a given time t, mark the first t corrupted positions, then count distinct substrings that include at least one '*' using a suffix automaton or suffix array with LCP, and compare to m. Binary search the smallest t where the count >= m, handling the initial check separately.
Pro tip: Clarify that 'distinct substrings' means distinct strings, not occurrences, and that the count can be computed as total distinct substrings minus distinct substrings with no '*'. This reduces the problem to counting substrings avoiding a set of forbidden positions, which can be done with a suffix automaton by resetting the last state at each forbidden character.
Confirm that 'unrecoverable' means the number of distinct substrings containing at least one '*' is >= m. Check if the initial string already meets this condition (return 1) and note that if m is 0, answer is 0.
Given t, mark the first t positions from the permutation as '*'. Count the number of distinct substrings that contain at least one '*'. This can be computed as total distinct substrings of the current string minus distinct substrings that contain no '*' (i.e., substrings entirely within segments of non-'*' characters).
Use a suffix automaton or suffix array to compute total distinct substrings. For substrings without '*', split the string by '*' and sum distinct substrings of each segment. Alternatively, build a suffix automaton and reset the last state at each '*' to count only substrings avoiding '*'.
Binary search t from 1 to n (or 0 to n-1 depending on indexing). For each mid, run the feasibility check. If count >= m, search left; else search right. Return the smallest t that satisfies the condition.
Each feasibility check takes O(n log n) or O(n) with suffix automaton. Binary search adds a log n factor, giving O(n log^2 n) or O(n log n). Discuss potential optimizations like incremental updates or using a segment tree to maintain counts.
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