My first instinct was to just dump everything into a set and sort it, which works but completely misses the point.
Start by clarifying the problem: the inputs are sorted, duplicate-free iterators, and the output should be a sorted iterator with no duplicates. Then describe a two-pointer merge approach that compares the current values from each iterator, advances the one with the smaller value, and when values are equal, advances both and emits one copy. Emphasize that the solution should be lazy (iterator-based) and handle edge cases like one iterator being exhausted.
Pro tip: Mention that this is essentially a merge step of merge sort with deduplication, and highlight that the iterator interface requires lazy evaluation to avoid materializing the entire input. Also, proactively discuss how you would test it with edge cases like empty iterators, one iterator being a subset of the other, and interleaved values.
Confirm that inputs are sorted and duplicate-free, output must be sorted and duplicate-free, and the solution should be lazy (iterator-based). Ask about the iterator interface (e.g., hasNext(), next()) and whether inputs can be empty.
Explain that you will maintain the current value from each iterator. At each step, compare the two values: if one is smaller, emit it and advance that iterator; if equal, emit one copy and advance both.
Describe how to handle when one iterator is exhausted: simply emit the remaining values from the other iterator. Also handle the case where both are exhausted by signaling the end of iteration.
State that time complexity is O(n + m) where n and m are the lengths of the input iterators, and space complexity is O(1) extra space (excluding the output). Mention that the lazy approach avoids storing all elements in memory.
Walk through a simple implementation in pseudocode or a language of choice, showing the main loop and the handling of equal values and exhaustion.
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