I got the stack approach pretty quickly: track each character alongside a count, and whenever a count hits 2 (or k in the follow-up), collapse it.
Use a stack to process the string character by character, merging with the top of the stack when a run of length >= 2 is formed, and then repeatedly removing any newly formed runs at the top. This yields an O(n) time and O(n) space solution. For the follow-up, adapt the stack to track run lengths and remove runs of length >= k.
Pro tip: Emphasize that the stack approach naturally handles cascading removals and is optimal; mention that a naive simulation would be O(n^2) due to repeated scans. For the follow-up, note that the same stack logic works with a threshold k, but careful implementation is needed to avoid missing cascades.
Restate the problem to ensure understanding: repeatedly remove maximal runs of length >= 2 until none exist. Discuss edge cases: empty string, no runs, entire string removed, and overlapping runs after concatenation.
Explain that a stack can simulate the process in one pass: push characters, and when the top forms a run of length >= 2, pop the entire run. This automatically handles concatenation and cascading removals.
Describe the algorithm step-by-step: iterate through the string, push each character onto the stack, and after each push, check if the top run has length >= 2; if so, pop it. Argue correctness by showing the stack maintains the invariant that it represents the reduced string after processing the prefix, and that any removal is applied immediately.
Each character is pushed and popped at most once, so time is O(n). The stack uses O(n) space in the worst case. This meets the linear-time, linear-space requirement.
Modify the stack to store characters along with their run lengths. When a run reaches length k, pop it. This still runs in O(n) time and O(n) space, as each character is processed once.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.