Google·Software Engineer·Onsite - System Design / Architecture·Senior

Senior

Jun 2026

Summary

Google SWE interview with a system design flavored coding question about a restaurant waitlist. More nuanced than a typical LeetCode problem since you had to think carefully about the tradeoffs between data structures rather than just grinding toward a solution.

Questions Asked (1)

Design a restaurant waitlist data structure supporting three operations: adding a user with a required party size, removing a user by name, and finding the first user in arrival order whose party size fits a given table capacity. The found user stays on the waitlist until explicitly removed. Walk through your data structure choices and the time complexity of each operation.

Algorithms & Data StructuresSystem DesignTechnical Trade-offs

Author's notes

I went with a doubly linked list to preserve arrival order plus a hashmap from user to node for O(1) deletes.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the requirements and constraints, then propose a data structure that balances the three operations. A common approach is to combine a hash map for O(1) removal by name with a balanced binary search tree (e.g., TreeMap) keyed by arrival order to efficiently find the first fitting party. Discuss trade-offs and possible optimizations.

Pro tip: Mention that the found user stays on the waitlist, so you must not remove them during the search; this implies the search operation should be read-only. Also, consider using a segment tree or Fenwick tree over party sizes to achieve O(log n) search if the waitlist is large.

1. Clarify requirements and constraints

Ask about expected number of operations, whether party sizes are bounded, and if the waitlist order is strictly by arrival time. Confirm that the found user remains on the waitlist.

2. Propose a hybrid data structure

Use a hash map (name -> node) for O(1) removal and a balanced BST (e.g., TreeMap) keyed by arrival order to maintain the queue. Each node stores party size.

3. Design the search operation

To find the first fitting party, traverse the BST in arrival order (in-order) and check party size <= capacity. This is O(n) worst-case, but can be optimized with a segment tree over party sizes to O(log n).

4. Analyze time complexities

Add: O(log n) for BST insertion + O(1) for hash map. Remove: O(log n) for BST deletion + O(1) for hash map. Search: O(n) with naive BST, O(log n) with segment tree.

5. Discuss trade-offs and optimizations

Compare naive BST vs. segment tree vs. bucket by party size. Mention that if party sizes are small, an array of queues per size can give O(1) search but may not preserve global arrival order.

Key Points to Mention

Hash map for O(1) removal by name, storing pointers to nodes in the order structure.
Balanced BST (e.g., TreeMap) or skip list to maintain arrival order and support O(log n) insertion/deletion.
Search operation must not remove the user; it should return the user but leave them in the waitlist.
Optimization: segment tree or Fenwick tree over party sizes to achieve O(log n) search for first fitting party.
Trade-offs: simplicity vs. performance; if party sizes are bounded, bucket by size with queues per size, but need to handle global order.
Edge cases: empty waitlist, no fitting party, duplicate names, and party size larger than any table.

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