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Early-stage Startup·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
Aug 2026Remote

Summary

Came out of a coding interview still a bit rattled by a graph/tree-looking problem that turned out to need a map of lists instead. Not the first time this pattern has tripped me up, and probably not the last.

Questions Asked (1)

Q1

Given a list of existing users and a parallel list where each index represents which existing user invited a new user, find the user who has the highest total invite count, where that count includes all direct and indirect invites down the chain.

Algorithms & Data Structures
Author's notes

My first instinct was to build a tree and do some kind of DFS, which is exactly what they want you to think.

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AI HintsAI Generated

Suggested Approach

Model the invite relationships as a directed graph or forest where each new user has exactly one parent (the inviter). Then compute the total number of descendants for each existing user, either via DFS with memoization or by processing nodes in reverse topological order, and return the user with the maximum count.

Pro tip: Clarify edge cases upfront—such as multiple roots, cycles, or users with zero invites—and mention that you'd validate the input to ensure it forms a valid forest. This shows you think about robustness beyond the happy path.

1. Clarify the problem and constraints

Ask about input size, whether the graph is guaranteed to be a forest (no cycles, each new user has exactly one inviter), and whether ties should be handled in a specific way. Confirm that 'total invite count' includes all descendants, not just direct invites.

2. Choose a representation

Build an adjacency list mapping each existing user to the list of users they directly invited. Alternatively, use a parent array where parent[i] is the inviter of new user i, then invert it to get children lists.

3. Select an algorithm

Use DFS with memoization to compute the size of the subtree rooted at each user, or process nodes in reverse topological order (e.g., via Kahn's algorithm) to accumulate counts bottom-up. Both run in O(N) time and space.

4. Compute and track the maximum

During the traversal, maintain a running maximum of the total descendant count and the corresponding user. Handle ties by returning any or all users with the maximum count, as specified.

5. Analyze complexity and test

State that the time and space complexity are O(N), where N is the total number of users. Walk through a small example to verify correctness, including edge cases like a single user or a chain.

Key Points to Mention

  • Graph modeling: treat the invite relationships as a directed forest where each node has at most one parent.
  • DFS with memoization or reverse topological order to compute subtree sizes efficiently.
  • Time and space complexity: O(N) for both, where N is the total number of users.
  • Handling edge cases: multiple roots, cycles (if possible), users with no invites, and ties.
  • Input validation: ensure the parallel lists are consistent and form a valid forest.
  • Scalability: the solution works for large inputs due to linear complexity.

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