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Given s of letters and special ops where a letter appends, '*' deletes the last char, '#' duplicates the entire current string, and '%' reverses it, compute the k-th character of the final result (or '.' if out of bounds). The core challenge is handling potentially exponential growth from '#' by tracking lengths and mapping k backwards through duplications/reversals/deletions instead of constructing the full string.
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Leetcode 3614. Process String with Special Operations II
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