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Google·Software Engineer·Onsite - Coding / Algorithms·Intermediate

IntermediatePrefer not to say
Jul 2026

Summary

Two coding problems at Google for a software engineer role. Both were algorithm-heavy and had a bunch of follow-ups that kept going deeper than I expected. Felt okay on the first one, less okay on the second.

Questions Asked (2)

Q1

Given a list of meeting time intervals, find the minimum number of rooms needed so all meetings can run simultaneously without conflict.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

I went straight for the min-heap approach because I'd seen something like it before.

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AI HintsAI Generated

Suggested Approach

Clarify the problem constraints and edge cases, then propose an efficient solution using a sweep line or priority queue approach. Explain the algorithm step-by-step, analyze time and space complexity, and discuss potential trade-offs.

Pro tip: Mention that sorting by start times and using a min-heap of end times is a common pattern for interval problems, and highlight that the number of rooms equals the maximum number of overlapping meetings at any point.

1. Clarify the problem

Ask about input format, whether intervals are inclusive/exclusive, if meetings can be back-to-back, and expected constraints (e.g., number of meetings).

2. Discuss approaches

Compare brute-force (check all pairs) with optimized methods like sweep line or min-heap. Explain why the optimized approach is better.

3. Detail the algorithm

Describe sorting intervals by start time, then using a min-heap to track end times. For each meeting, if the earliest end time is <= current start, reuse a room; otherwise allocate a new one.

4. Analyze complexity

State that sorting takes O(n log n) and heap operations take O(n log n), resulting in O(n log n) time and O(n) space.

5. Test with examples

Walk through a small example to verify correctness and edge cases like empty input or all meetings overlapping.

Key Points to Mention

  • Sorting intervals by start time
  • Using a min-heap to track end times of ongoing meetings
  • The number of rooms equals the maximum number of concurrent meetings
  • Time complexity: O(n log n) due to sorting and heap operations
  • Space complexity: O(n) for the heap
  • Handling edge cases: empty input, back-to-back meetings, and overlapping intervals

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

Q2

Design a component that maintains the top K items by score from a stream of (itemId, score) pairs, with tie-breaking by smaller itemId.

Algorithms & Data StructuresSystem DesignTechnical Trade-offs
Author's notes

This one spiraled.

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AI HintsAI Generated

Suggested Approach

Start by clarifying requirements: stream is unbounded, K is fixed, need top K by score with tie-breaking by smaller itemId. Then propose a min-heap of size K where the heap ordering is by (score, -itemId) so the root is the worst among the top K, allowing O(log K) updates per item. Discuss trade-offs with alternative approaches like balanced BST or sorted list, and handle edge cases like duplicate itemIds and K=0.

Pro tip: Mention that you can optimize for the common case where the new item's score is less than the heap root's score by doing an O(1) comparison before any heap operation, and discuss how to handle updates to existing items if the stream allows duplicates.

1. Clarify requirements and constraints

Ask about stream characteristics (unbounded, online), K size, whether itemIds can repeat, and if updates to existing items are allowed. Confirm tie-breaking rule: higher score first, then smaller itemId.

2. Choose data structure and define ordering

Propose a min-heap of size K. Define the heap comparator: items with lower score are 'smaller'; for equal scores, the item with larger itemId is 'smaller' (so the root is the worst among top K).

3. Detail insertion and eviction logic

For each new (itemId, score): if heap size < K, push; else if (score, -itemId) > (root.score, -root.itemId), pop root and push new item. Otherwise, ignore. This maintains top K.

4. Analyze complexity and trade-offs

Time: O(log K) per item worst-case, O(1) for rejected items. Space: O(K). Compare with alternatives: balanced BST (O(log K) but higher constants), sorted array (O(K) insertion), or keeping all items (O(N) space).

5. Handle edge cases and extensions

Discuss K=0, duplicate itemIds (if updates allowed, need a map from itemId to heap node for O(log K) update), and potential concurrency if stream is parallel. Mention that if K is large, consider a more advanced structure like a Fibonacci heap or a skip list.

Key Points to Mention

  • Min-heap of size K with custom comparator: (score, -itemId) to ensure correct tie-breaking.
  • O(log K) insertion and O(1) rejection for items that cannot be in top K.
  • Space complexity O(K), which is optimal for this problem.
  • Trade-offs: heap vs. balanced BST vs. sorted list; heap is simpler and has better constants for this use case.
  • Handling duplicate itemIds: if updates are allowed, maintain a hash map from itemId to heap index for O(log K) updates.
  • Edge cases: K=0, empty stream, and items with equal scores and itemIds (if duplicates allowed).

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