Clarify the iterator interface and constraints, then propose a two-pointer merge that advances the iterator with the smaller current value and skips duplicates. Discuss edge cases like empty iterators and infinite streams, and analyze time and space complexity.
Pro tip: Emphasize that the solution should be lazy and streaming, handling infinite sequences without loading all data into memory, and mention that the same logic can be adapted for intersection or difference.
Ask about the iterator interface (e.g., hasNext/next), whether inputs are truly sorted and deduplicated, and if they can be infinite. Confirm output should be a new iterator, not a list.
Explain that you will maintain the current value from each iterator, compare them, and emit the smaller one while advancing that iterator. If equal, emit once and advance both.
Describe how to skip consecutive duplicates within each iterator (if not guaranteed deduplicated) and handle empty iterators or one iterator exhausting before the other.
Show how to implement hasNext() and next() without precomputing the entire union, using a state machine or buffering the next value. Ensure next() only advances when called.
State that time complexity is O(n+m) total across all next() calls, and space is O(1) extra. Discuss trade-offs: lazy vs eager, and how the approach extends to intersection or difference.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by restating the problem and clarifying constraints (e.g., k can be large, iterators may be slow, duplicates already removed per iterator). Then propose a min-heap of size k to efficiently merge the iterators, and analyze time and space complexity. Finally, discuss trade-offs and potential optimizations like using a loser tree or tournament tree for large k.
Pro tip: Mention that the heap stores the current element from each iterator along with the iterator index, and that you must handle the case where an iterator is exhausted. Also, note that if k is very large, a heap may not be optimal and a tournament tree could reduce comparisons.
Confirm that each iterator is sorted and internally deduplicated, and that we need to merge them into a single sorted stream without duplicates. Ask about k's typical size and whether the output should be an iterator or a list.
Propose a min-heap (priority queue) of size k, where each entry contains the current value and the iterator index. Explain that this allows efficient retrieval of the next smallest element across all iterators.
Describe initializing the heap with the first element from each non-empty iterator. Then repeatedly extract the minimum, output it, and push the next element from the same iterator if available. Also, skip duplicates if the same value appears from multiple iterators.
Time: O(N log k) where N is total number of elements, because each element is pushed and popped once, each operation O(log k). Space: O(k) for the heap, plus O(1) extra if output is streamed.
Mention that for very large k, a heap may have high constant factors; a tournament tree (loser tree) can reduce comparisons to O(log k) but with lower constants. Also, if k is small, a simple linear scan might be faster.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.