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Snowflake·Software Engineer·Technical Phone Screen·Senior

SeniorPrefer not to say
May 2026Remote

Summary

Snowflake SWE interview that went deep into graph algorithms pretty fast. The question looked like a clean CS problem on the surface but they really wanted to see if you understood the performance implications at scale, which is very on-brand for a data warehouse company.

Questions Asked (1)

Q1

Given a directed acyclic graph, compute the transitive closure so that for each node you know every other node reachable from it. Then discuss the trade-offs between doing a DFS from each node versus a topological-order propagation approach using bitsets.

Algorithms & Data StructuresTechnical Trade-offsSystem Design
Author's notes

I started with the naive DFS-per-node answer because that's the obvious one and I figured I'd build from there.

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AI HintsAI Generated

Suggested Approach

First, clearly define the problem and the output format (e.g., a bitset per node). Then present two solutions: (1) DFS from each node with memoization to avoid redundant work, and (2) topological-order propagation using bitsets. Compare their time and space complexities, and discuss practical trade-offs such as graph density, memory constraints, and implementation complexity.

Pro tip: Mention that in practice, the bitset approach is often faster for dense graphs due to cache-friendly bitwise operations, but for sparse graphs or when memory is tight, DFS with memoization can be more efficient. Also, note that the bitset approach requires O(V^2) memory, which may be prohibitive for large graphs.

1. Clarify the problem and constraints

Confirm the input format (adjacency list or matrix), output format (e.g., boolean matrix or bitsets), and any constraints on graph size, density, or memory. Ask if the graph is static or dynamic.

2. Describe the DFS approach

Explain that you can run DFS from each node, using memoization to store reachable sets and avoid recomputation. Analyze time complexity: O(V*(V+E)) without memoization, but with memoization it can be O(V*E) or better depending on sharing.

3. Describe the topological-order bitset approach

Process nodes in reverse topological order. For each node, initialize a bitset with itself, then OR the bitsets of all its successors. This yields the transitive closure. Time complexity: O(V*E/word_size) due to bitwise operations, space O(V^2/word_size).

4. Compare trade-offs

Discuss time and space complexity, practical performance (cache efficiency, constant factors), and suitability for different graph densities. Mention that bitset approach is often faster for dense graphs but uses more memory; DFS with memoization can be more memory-efficient for sparse graphs.

5. Conclude with a recommendation

Summarize which approach you would choose based on typical constraints (e.g., if V is up to a few thousand, bitset is fine; if V is huge and graph sparse, DFS with memoization). Mention that both can be optimized further (e.g., using bitsets in DFS).

Key Points to Mention

  • Time complexity of DFS from each node: O(V*(V+E)) naive, O(V*E) with memoization.
  • Time complexity of topological bitset propagation: O(V*E/word_size) due to bitwise OR operations.
  • Space complexity: DFS with memoization can be O(V^2) in worst case if storing all reachable sets, but can be optimized; bitset approach requires O(V^2/word_size) memory.
  • Bitset approach leverages bit-level parallelism and is cache-friendly, often faster in practice for dense graphs.
  • DFS with memoization can share reachable sets and avoid redundant work, beneficial for sparse graphs or when memory is limited.
  • Topological order ensures each node is processed after all its successors, enabling efficient propagation.

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