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Snowflake·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Jul 2026

Summary

Snowflake software engineer interview with a graph traversal problem. Pretty standard BFS setup but the edge cases tripped me up more than I expected.

Questions Asked (1)

Q1

Given a start and target wiki page URI, implement a function that returns the minimum number of clicks (link hops) to get from one to the other. You have access to an API that returns all outgoing links from a given page. Return 0 if they're the same page, -1 if unreachable.

Algorithms & Data StructuresAPI & Integrations
Author's notes

My first instinct was DFS and I'm glad I caught myself before coding it up because that would've been wrong.

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AI HintsAI Generated

Suggested Approach

Model the wiki pages as a graph and use BFS to find the shortest path in terms of link hops. Start from the source page, explore level by level, and return the distance when the target is found; if the queue empties, return -1. Handle the trivial case where start equals target by returning 0 immediately.

Pro tip: Mention that bidirectional BFS can significantly reduce the search space, especially for distant pages, and discuss how to handle API latency with caching or concurrent requests.

1. Clarify requirements and edge cases

Confirm that the API returns outgoing links and that we need the minimum number of hops. Discuss edge cases: same page (return 0), unreachable (return -1), and potential cycles.

2. Choose BFS for shortest path

Explain that BFS guarantees the shortest path in an unweighted graph. Use a queue to track pages to visit and a set to track visited pages to avoid cycles.

3. Implement BFS with level tracking

Initialize queue with start page and distance 0. While queue not empty, dequeue a page and its distance; if it's the target, return distance. Otherwise, fetch its outgoing links, and for each unvisited link, mark visited and enqueue with distance+1.

4. Optimize with bidirectional BFS

If performance is a concern, propose bidirectional BFS: run two BFS from start and target simultaneously, expanding the smaller frontier, and stop when they meet. This reduces time and space complexity.

5. Analyze complexity and discuss trade-offs

State time complexity O(V+E) where V is pages visited and E is links explored, and space O(V). Mention that bidirectional BFS can reduce the branching factor, and discuss caching API results to avoid repeated calls.

Key Points to Mention

  • BFS is optimal for unweighted shortest path problems.
  • Use a visited set to prevent infinite loops due to cycles.
  • Handle the trivial case where start equals target by returning 0.
  • Return -1 if BFS exhausts all reachable pages without finding the target.
  • Consider bidirectional BFS to improve efficiency on large graphs.
  • Cache API responses to reduce latency and avoid redundant calls.

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