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Google·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
Jun 2026Remote

Summary

Google technical phone screen, graph problem that started straightforward and then got progressively more interesting once the follow-up kicked in. Felt okay about it but the union-find discussion at the end was where I was least confident.

Questions Asked (2)

Q1

You're given a list of nodes, each with a primary ID and a secondary attribute like a color or label. Two nodes are connected if they share that attribute. Build an adjacency list and run BFS from a source node to answer reachability or shortest-path queries.

Algorithms & Data StructuresSystem Design
Author's notes

Started with the naive O(n^2) pairwise comparison and they let me run with it for a bit before asking if I could do better.

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AI HintsAI Generated

Suggested Approach

Clarify the problem constraints and define the graph model: nodes are connected if they share the secondary attribute. Build an adjacency list by grouping nodes by attribute, then run BFS from the source to answer reachability or shortest-path queries. Discuss time/space complexity and potential optimizations.

Pro tip: Mention that building the adjacency list can be done in O(N) time by using a hash map from attribute to list of nodes, and that BFS is optimal for unweighted graphs. Also, consider if the graph is static or dynamic, as that affects whether to precompute or compute on the fly.

1. Clarify the problem

Ask about constraints: number of nodes, attributes, query types (reachability vs shortest path), and whether the graph is static or dynamic. Confirm that edges are undirected and unweighted.

2. Model the graph

Represent each node with its primary ID and secondary attribute. Group nodes by attribute using a hash map to efficiently build the adjacency list.

3. Build adjacency list

For each attribute, connect all nodes sharing that attribute. This can be done by iterating through the attribute groups and adding edges between every pair of nodes in the group.

4. Run BFS

Perform BFS from the source node to compute distances or reachability. Use a queue and a visited set to avoid cycles.

5. Analyze and optimize

Discuss time and space complexity: O(N + E) for BFS, where E can be O(N^2) in worst case. Consider optimizations like early termination for reachability queries or precomputing connected components if many queries.

Key Points to Mention

  • Time complexity: O(N + E) for BFS, where E is number of edges; building adjacency list can be O(N + total pairs per attribute).
  • Space complexity: O(N + E) for adjacency list and BFS structures.
  • Use of hash map to group nodes by attribute for efficient adjacency list construction.
  • BFS guarantees shortest path in unweighted graphs.
  • Handling disconnected graphs and unreachable nodes.
  • Potential optimizations: precompute connected components if many queries, or use union-find for reachability only.

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

Q2

Follow-up: when would you use union-find instead of the attribute-keyed map approach, and what are the tradeoffs?

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

This is where I got a bit shaky.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem context—specifically whether you need to track connected components dynamically or just group items by attributes. Then compare union-find and attribute-keyed maps on operations (union, find, iteration), complexity, and memory, and conclude with concrete scenarios where each shines.

Pro tip: Mention that union-find with path compression and union by rank gives near-constant amortized time per operation, but it doesn't support efficient enumeration of all members in a component—a key limitation that attribute-keyed maps handle naturally.

1. Clarify the problem requirements

Ask whether the problem involves dynamic connectivity, repeated unions, or just static grouping. This determines if union-find is even applicable.

2. Compare core operations

Contrast union-find's union/find (near O(1) amortized) with attribute-keyed maps' O(1) insert/lookup and O(k) iteration over a group. Highlight that union-find cannot efficiently list all elements in a component.

3. Analyze tradeoffs

Discuss memory overhead (union-find uses parent/rank arrays; maps use hash tables), flexibility (maps allow multiple attributes and easy updates; union-find only tracks connectivity), and complexity of implementation.

4. Give concrete use cases

Provide examples: union-find for Kruskal's MST, dynamic connectivity in networks, or image segmentation; attribute-keyed maps for grouping users by role, indexing by multiple keys, or when frequent iteration over groups is needed.

5. Conclude with a recommendation

Summarize that union-find is ideal for dynamic connectivity with many unions and few queries, while attribute-keyed maps are better for static grouping, multi-attribute indexing, and when you need to enumerate group members.

Key Points to Mention

  • Union-find excels at dynamic connectivity problems with many union operations and few find queries.
  • Attribute-keyed maps allow O(1) access to groups by key and efficient iteration over all members of a group.
  • Union-find with path compression and union by rank achieves near-constant amortized time per operation.
  • Union-find does not support efficient enumeration of all elements in a component without additional data structures.
  • Attribute-keyed maps can handle multiple attributes and are more flexible for complex grouping and filtering.
  • Memory overhead: union-find uses arrays of size N; maps use hash tables with potential overhead for keys and values.

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