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Google·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Apr 2026

Summary

Google SWE interview with a queue-based algorithmic problem that sounds deceptively clean on the surface but gets tricky fast once you start thinking about minimizing operations. The core challenge was figuring out how to interleave polling across multiple queues efficiently rather than just draining them one at a time.

Questions Asked (1)

Q1

You have a collection of queues where you can only call isEmpty() or poll() (no peeking). Find the minimum number of elements across all queues and the minimum sum of elements across all queues, while minimizing the total number of isEmpty/poll calls. How do you interleave polling so you stop as early as possible and avoid unnecessary draining?

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

This one took me a minute to even understand what the actual optimization target was.

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AI HintsAI Generated

Suggested Approach

Model the problem as finding the minimum length and sum across queues by polling elements in rounds, stopping as soon as any queue becomes empty. Use a round-robin polling strategy that checks isEmpty() before each poll, and maintain running totals to avoid unnecessary polls once a queue is exhausted.

Pro tip: Emphasize that the optimal strategy is to poll one element from each queue per round, because the minimum length is determined by the shortest queue; once any queue is empty, you can stop polling all queues. This minimizes total calls by avoiding draining longer queues.

1. Understand the constraints and goal

Recognize that you can only call isEmpty() and poll(), and you need to find the minimum number of elements and minimum sum across all queues while minimizing total calls. The minimum length is the length of the shortest queue, and the minimum sum is the sum of the first minLength elements of each queue.

2. Design a round-robin polling strategy

Poll one element from each queue in each round, checking isEmpty() before polling. Keep a running sum for each queue and a count of elements polled. Stop as soon as any queue returns true for isEmpty() after polling, because that queue is now empty and its length is the minimum.

3. Track minimum length and sum

Maintain a global count of rounds completed (which equals the number of elements polled from each non-empty queue) and a global sum of all polled elements. When a queue becomes empty, the current round count is the minimum length, and the global sum is the minimum sum.

4. Handle edge cases and optimize calls

If a queue is initially empty, the minimum length is 0 and minimum sum is 0, requiring only one isEmpty() call per queue. Also, avoid polling from queues that are already known to be empty in subsequent rounds.

5. Analyze complexity and trade-offs

The total number of calls is O(k * minLength) where k is the number of queues, which is optimal because you must poll at least minLength elements from each queue to compute the sum. Discuss that this is better than draining all queues, which would be O(total elements).

Key Points to Mention

  • The minimum number of elements across all queues is the length of the shortest queue.
  • The minimum sum is the sum of the first minLength elements of each queue.
  • Round-robin polling ensures you stop as soon as the shortest queue is exhausted.
  • Total calls are minimized by not polling from queues after they are known to be empty.
  • Edge case: if any queue is initially empty, answer is 0 with minimal calls.
  • Time complexity is O(k * minLength), which is optimal for this problem.

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