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Oracle·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Jun 2026

Summary

Oracle SWE interview with a data structures question that sounds straightforward until you actually have to think through the edge cases under pressure. One question, design-focused, with a discussion component about trade-offs.

Questions Asked (1)

Q1

Given a continuous stream of user login events, design a data structure that supports two O(1) operations: one to record that a user has logged in (users can log in multiple times), and one to return the earliest-logged-in user who has logged in exactly once. Return null if no such user exists.

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

I knew a linked list was involved pretty early but kept second-guessing myself on how to handle the duplicates cleanly.

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AI HintsAI Generated

Suggested Approach

Clarify the requirements and constraints, then propose a design using a hash map to track login counts and a doubly linked list to maintain the order of users who have logged in exactly once. Explain how to achieve O(1) for both operations by updating the list and map on each login, and discuss edge cases and trade-offs.

Pro tip: Mention that the solution must handle multiple logins per user and that the order is based on the first login time among users with exactly one login. Also, discuss how to handle deletions from the linked list efficiently using a hash map to store node references.

1. Clarify Requirements

Ask clarifying questions to confirm assumptions: e.g., whether 'earliest-logged-in' refers to the first login event among users who have logged in exactly once, and whether the stream is continuous with no deletions.

2. Choose Data Structures

Select a hash map to store each user's login count and a doubly linked list to maintain the order of users with exactly one login. Also, use a second hash map to store references to nodes in the linked list for O(1) removal.

3. Define Operations

For recordLogin(user): increment the user's count; if count becomes 1, append the user to the linked list; if count becomes 2, remove the user from the linked list. For getEarliestSingleLogin(): return the head of the linked list or null if empty.

4. Analyze Complexity

Explain that both operations run in O(1) time because hash map lookups and linked list insertions/removals are constant time. Space complexity is O(n) for n unique users.

5. Discuss Edge Cases and Trade-offs

Cover scenarios like no users with exactly one login, multiple logins by the same user, and potential memory overhead. Mention alternative approaches (e.g., using a queue with lazy deletion) and their trade-offs.

Key Points to Mention

  • Use a hash map to track login counts per user.
  • Maintain a doubly linked list to preserve insertion order of users with exactly one login.
  • Store node references in a hash map for O(1) removal from the linked list.
  • Update the linked list only when a user's login count transitions between 1 and 2.
  • Both operations achieve O(1) time complexity.
  • Handle edge cases: empty list, user logging in multiple times, and memory considerations.

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