MongoDB·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Jun 2026Remote

Summary

MongoDB SWE interview with a single coding question about merging two sorted iterators. Pretty focused session, no fluff, just the problem and a lot of edge case poking.

Questions Asked (1)

Given two iterators over pre-sorted sequences, implement a UnionIterator class with hasNext() and getNext() that returns elements in sorted order with duplicates removed across both sources. You cannot load everything into memory; you must use O(1) extra space beyond the iterators themselves.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

I got the general shape pretty fast but the deduplication tripped me up more than I expected.

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AI HintsAI Generated

Suggested Approach

Use a k-way merge approach with two iterators, maintaining the current element from each iterator and advancing the one with the smaller value. To handle duplicates, skip any element that equals the last returned value, ensuring each unique element is emitted once. This uses O(1) extra space and leverages the pre-sorted nature of the inputs.

Pro tip: Clarify the iterator interface upfront (e.g., hasNext/getNext) and handle edge cases like empty iterators and duplicate values at the boundaries. Mention that the solution is generic and can be extended to k iterators with a heap, but for two iterators, a simple comparison suffices.

1. Understand the problem and constraints

Restate the problem: merge two sorted iterators, remove duplicates, O(1) extra space. Confirm the iterator interface and that inputs are sorted.

2. Design the algorithm

Maintain the current element from each iterator. At each step, compare the two current elements, pick the smaller one, and advance that iterator. Skip if the picked element equals the last returned value.

3. Handle edge cases

Consider empty iterators, one iterator exhausted, and duplicates across iterators. Ensure hasNext correctly reflects whether any unique elements remain.

4. Implement and test

Write clean code with clear variable names. Test with cases like [1,2,3] and [2,3,4], [1,1,1] and [1,1], and empty inputs.

5. Analyze complexity and trade-offs

Explain O(1) space and O(n+m) time. Discuss alternative approaches (e.g., using a heap for k iterators) and why they are not needed here.

Key Points to Mention

O(1) extra space: only store current elements and last returned value, no additional data structures.
Duplicate removal: compare with last returned value and skip if equal.
Iterator advancement: only advance the iterator whose current element is chosen.
Edge cases: empty iterators, one exhausted, all duplicates.
Time complexity: O(n+m) where n and m are the lengths of the sequences.
Generality: the approach can be extended to k sorted iterators using a min-heap, but for two, simple comparison is optimal.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.