Started with the brute force and re-scanned the whole sequence after every removal.
Model the sequence as a doubly linked list to enable O(1) removal and neighbor access. Use a queue to process eligible items in order, checking only the two neighbors after each removal and enqueueing them if they become eligible. This yields O(N) time and O(N) space, avoiding the naive O(N^2) repeated scans.
Pro tip: Clarify upfront whether the predicate is static or dynamic; if dynamic, discuss how to handle re-evaluation. Also, mention edge cases like empty input, all eligible, or none eligible to show thoroughness.
Ask about input size, whether the predicate can change, and if the output should be the remaining sequence or just the count. Confirm that only immediate neighbors can become eligible after a removal.
Use a doubly linked list to represent the sequence for O(1) removals and neighbor access. Use a queue (or stack) to process eligible items in the order they become eligible.
Initialize the linked list and enqueue all initially eligible items. While the queue is not empty, dequeue an item, remove it from the list, and check its left and right neighbors; if they become eligible, enqueue them.
Explain that each item is enqueued at most once, so total operations are O(N). Compare with naive O(N^2) and discuss space trade-off (O(N) extra for queue and linked list pointers).
Consider empty input, all items eligible, none eligible, and cases where removals cascade. Walk through a small example to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.