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Snowflake·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Jun 2026

Summary

Snowflake software engineer interview with a grid pathfinding problem and a follow-up on DFS vs BFS tradeoffs. Pretty standard algorithmic round but the follow-up is where things got interesting.

Questions Asked (2)

Q1

Given an m x n grid of open and blocked cells, find the minimum number of moves to get from a start cell to a target cell moving in four directions. Return -1 if unreachable.

Algorithms & Data Structures
Author's notes

Classic BFS setup, went fine.

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AI HintsAI Generated

Suggested Approach

Model the grid as a graph where each open cell is a node connected to its four neighbors. Use BFS from the start cell to find the shortest path to the target, tracking the number of moves. If the target is never reached, return -1.

Pro tip: Clarify edge cases upfront: what if start or target is blocked? What if they are the same cell? Handling these gracefully shows attention to detail. Also, mention that BFS is optimal for unweighted grids, but if the grid is huge, consider bidirectional BFS to reduce search space.

1. Understand the problem and constraints

Confirm grid dimensions, movement rules (4-directional), and what constitutes a valid move. Ask about edge cases like blocked start/target or same start and target.

2. Choose the right algorithm

Recognize that this is a shortest path problem on an unweighted graph, so BFS is ideal. Mention that DFS would not guarantee shortest path.

3. Implement BFS with a queue

Initialize a queue with the start cell and a visited set or distance matrix. Process cells level by level, exploring neighbors in all four directions.

4. Track distance and handle termination

Increment distance as you move to the next level. Return the distance when the target is reached; if the queue empties, return -1.

5. Analyze complexity and optimize

State time and space complexity: O(m*n) for both. Discuss potential optimizations like bidirectional BFS or early exit if target is found.

Key Points to Mention

  • BFS guarantees shortest path in unweighted graphs
  • Use a queue for level-order traversal
  • Track visited cells to avoid cycles and redundant work
  • Handle edge cases: start/target blocked, same cell, out-of-bounds
  • Time and space complexity: O(m*n)
  • Potential optimization: bidirectional BFS for large grids

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

Q2

How would you solve the same grid pathfinding problem using recursive DFS with backtracking and memoization, and what are the tradeoffs versus BFS?

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

This is the part I was less prepared for.

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AI HintsAI Generated

Suggested Approach

Start by clearly defining the recursive DFS with backtracking and memoization: a function that explores all valid moves from the current cell, marks the cell as visited, recurses, then unmarks it. Use memoization to cache results for each cell to avoid redundant computations. Then compare with BFS, focusing on tradeoffs in time/space complexity, optimality, and practical use cases.

Pro tip: Emphasize that memoization in DFS is only valid if the problem has optimal substructure and overlapping subproblems; otherwise, it can lead to incorrect results. Also, mention that BFS is generally preferred for shortest path in unweighted grids, while DFS with memoization is better for counting paths or when memory is constrained.

1. Explain the recursive DFS approach

Describe the recursive function that takes the current position and explores all valid neighboring cells, using a visited set to avoid cycles. After exploring, backtrack by unmarking the cell.

2. Incorporate memoization

Add a memo table (e.g., 2D array) to store results for each cell, such as the number of paths from that cell to the target. Before recursing, check if the result is already computed; if so, return it.

3. Analyze time and space complexity

State that with memoization, each cell is computed once, leading to O(m*n) time and space. Without memoization, it's exponential. Compare with BFS which is also O(m*n) time but uses O(m*n) space for the queue.

4. Compare tradeoffs with BFS

Discuss that BFS guarantees shortest path in unweighted graphs and is iterative (no recursion depth issues), while DFS with memoization can be more memory-efficient for certain problems (e.g., counting paths) and avoids storing all frontier nodes.

5. Conclude with appropriate use cases

Summarize when to choose each: BFS for shortest path, DFS with memoization for counting paths or when the search space is deep and narrow, and mention that both can be adapted for weighted graphs with modifications.

Key Points to Mention

  • Recursive DFS with backtracking: mark visited, recurse, unmark.
  • Memoization: cache results per cell to avoid recomputation, ensuring O(m*n) time.
  • BFS guarantees shortest path in unweighted grids; DFS does not unless exhaustive search.
  • Space complexity: BFS uses queue (O(m*n)), DFS uses recursion stack (O(m*n) worst case) plus memo table.
  • DFS with memoization is ideal for counting paths or when memory is limited; BFS is better for shortest path.
  • Potential pitfalls: memoization invalid if state includes path-dependent information; recursion depth may cause stack overflow.

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