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Decagon·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
Jul 2026Remote

Summary

Decagon software engineer interview with a system design coding problem centered on a sliding window data structure. The follow-up pushed into update semantics which is where things got more interesting.

Questions Asked (2)

Q1

Design a class with an insert method and a get_avg method, where get_avg returns the average of all values whose timestamps fall within a fixed time window ending at the given timestamp. Timestamps are guaranteed to be non-decreasing across calls.

Algorithms & Data StructuresSystem Design
Author's notes

The monotonic timestamp constraint is the key thing to notice early.

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AI HintsAI Generated

Suggested Approach

Use a queue (or deque) to store (timestamp, value) pairs, and maintain a running sum of values in the window. On insert, add the new pair and update the sum; then remove from the front all pairs with timestamp < current_timestamp - window_size, subtracting their values from the sum. For get_avg, return sum / queue length (or 0 if empty).

Pro tip: Clarify whether the window is inclusive of the current timestamp and whether get_avg should be called with the same timestamp as the last insert. Also, mention that the non-decreasing timestamps guarantee the queue remains sorted, so we can efficiently evict old entries.

1. Clarify requirements and edge cases

Ask about window inclusivity, behavior when no values are in the window, and whether get_avg can be called with a timestamp earlier than the last insert. Confirm that timestamps are non-decreasing.

2. Choose data structures

Select a queue (e.g., collections.deque in Python) to store (timestamp, value) pairs in insertion order, and maintain a running sum variable to avoid recomputing the sum each time.

3. Implement insert

Append the new (timestamp, value) to the queue and add value to the running sum. Then, while the queue is not empty and the front timestamp is outside the window (i.e., < current_timestamp - window_size), remove it and subtract its value from the sum.

4. Implement get_avg

If the queue is empty, return 0 (or appropriate sentinel). Otherwise, return the running sum divided by the number of elements in the queue.

5. Analyze complexity and test

Explain that each element is inserted and removed at most once, so amortized O(1) time per operation and O(n) space. Walk through an example with multiple inserts and get_avg calls to verify correctness.

Key Points to Mention

  • Use a queue to maintain order and a running sum for O(1) average calculation.
  • Evict expired entries from the front of the queue when their timestamp falls outside the window.
  • Non-decreasing timestamps ensure that expired entries are always at the front, so no search is needed.
  • Handle edge cases: empty window (return 0 or None), window size zero, and timestamps exactly at the boundary.
  • Time complexity: amortized O(1) per insert and get_avg; space complexity O(n) where n is the number of elements in the window.
  • Consider thread-safety if the class might be used in a concurrent environment (optional, but shows maturity).

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

Q2

Follow-up: add an update method that can modify the value or refresh the timestamp of an existing record by its ID. What data structures would you need, and what are the tradeoffs between O(log n) and O(1) update complexity?

Algorithms & Data StructuresTechnical Trade-offsSystem Design
Author's notes

This is where I started fumbling a bit.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the requirements: update method should modify value or refresh timestamp by ID. Then propose a data structure that supports efficient lookup and update, such as a hash map for O(1) average update, or a balanced BST for O(log n) worst-case. Discuss tradeoffs including time complexity, memory overhead, and concurrency considerations.

Pro tip: Mention that O(1) hash map updates degrade to O(n) in worst-case due to collisions, while O(log n) structures like balanced BSTs provide predictable performance and ordered traversal, which can be valuable for range queries or timestamp ordering.

1. Clarify requirements

Confirm what 'update' entails: modifying value, refreshing timestamp, or both. Ask about expected read/write patterns and concurrency needs.

2. Propose data structures

Suggest a hash map (ID -> record) for O(1) average update, or a balanced BST (e.g., red-black tree) keyed by ID for O(log n) update. Mention hybrid approaches like hash map + linked list for LRU.

3. Analyze tradeoffs

Compare O(1) vs O(log n): hash map offers faster average updates but no ordering and worst-case O(n); BST provides ordered operations and stable O(log n) but higher constant factors.

4. Consider practical factors

Discuss memory overhead, cache performance, concurrency (locking vs lock-free), and whether ordering by timestamp is needed for queries.

5. Conclude with recommendation

Choose based on use case: if updates dominate and order isn't needed, hash map; if range queries or ordered traversal matter, BST.

Key Points to Mention

  • Hash map provides O(1) average update but O(n) worst-case due to collisions; balanced BST gives O(log n) worst-case.
  • Ordered operations (e.g., range queries by timestamp) are efficient with BST but not with hash map.
  • Memory overhead: hash map may have load factor and resizing overhead; BST has pointer overhead per node.
  • Concurrency: hash map can use fine-grained locking or lock-free techniques; BST updates may require rebalancing and more complex locking.
  • Timestamp refresh might require updating an index if queries rely on timestamp ordering.
  • Hybrid structures (e.g., hash map + skip list) can combine fast lookup with ordered access.

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