My first instinct was a brute force double loop and I almost said it out loud before catching myself.
Clarify the problem constraints (e.g., sorted input, definition of K-day window) and then propose an efficient solution using a hash map to track the last watched day for each episode. Iterate through the watch history, and for each episode, check if the difference between the current day and the last watched day is less than or equal to K; if so, return True, otherwise update the last watched day. This yields O(n) time and O(m) space, where n is the number of records and m is the number of unique episodes.
Pro tip: Mention that if the input is not sorted by day, you should sort it first (or use a sliding window with a set) to ensure the K-day window is correctly evaluated. Also, clarify whether the window is inclusive of both endpoints (i.e., difference <= K) to avoid off-by-one errors.
Ask about input size, whether the watch history is sorted by day, and the exact definition of 'within K days' (inclusive or exclusive). Confirm the expected return type.
Use a hash map (dictionary) to store the most recent day each episode was watched. This allows O(1) lookups and updates.
Traverse the watch history in chronological order. For each (episode_id, day), if the episode exists in the map and day - last_day <= K, return True. Otherwise, update the map with the current day.
If the loop completes without finding a re-watch, return False. Consider edge cases like empty input, K=0, or duplicate days.
State that the solution runs in O(n) time and O(m) space. If the input is unsorted, mention that sorting takes O(n log n) or that a sliding window with a set can be used.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.