Use a hash set to track the last k elements in the sliding window, adding each new element and removing the element that falls out of the window. If the new element is already in the set, return true; otherwise, after processing all elements, return false. This achieves O(n) time and O(k) space.
Pro tip: Clarify edge cases upfront, such as k <= 0 or k > stream length, and discuss how you would handle them. Also, mention that the solution can be adapted for streaming data with limited memory by using a Bloom filter if exactness is not required.
Restate the problem to ensure clarity: given a stream of integers and a window size k, determine if any duplicate appears within any window of size k. Confirm edge cases like k=1 (always false) or k=0 (invalid).
Select a hash set to store the current window's elements for O(1) average-time insertions, deletions, and lookups. Optionally, use a queue to maintain the order of elements for removal.
Iterate through the stream: for each element, check if it's in the set. If yes, return true. Otherwise, add it to the set and, if the set size exceeds k, remove the oldest element (using a queue or by tracking indices).
State that the time complexity is O(n) since each element is processed once, and space complexity is O(k) for the set and queue. Mention that this is optimal for the problem.
Walk through a small example, such as stream = [1,2,3,1], k=3, to demonstrate the algorithm returns true. Also test edge cases like no duplicates or k larger than stream length.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.