Snowflake·Software Engineer·Technical Phone Screen·Senior

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Apr 2026Remote

Summary

Snowflake coding interview with a tree pruning problem that looked straightforward until the follow-up on complexity hit. Not the worst session I've had but definitely left me second-guessing my DP formulation.

Questions Asked (1)

Given a rooted tree and an integer N, find the minimum number of nodes to delete so that the tree's maximum depth does not exceed N. When a node is deleted, its entire subtree is removed. Return the minimum number of deletions and discuss the complexity of your approach.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

I went with DFS to compute subtree depths first, which felt right, but then I stumbled when trying to articulate the greedy vs DP angle.

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AI HintsAI Generated

Suggested Approach

Clarify the problem: we need to delete the minimum number of nodes (each deletion removes the entire subtree) so that the resulting tree's maximum depth is at most N. Use a greedy bottom-up approach: for each node, compute the height of its subtree; if the height exceeds N, delete the node and increment the deletion count. This yields an optimal solution in O(V) time.

Pro tip: Emphasize that deleting a node is always at least as good as deleting any of its descendants when the subtree height exceeds N, because it removes more nodes and reduces depth more effectively. This greedy choice is optimal and can be proven by an exchange argument.

1. Clarify the problem and constraints

Confirm that the tree is rooted, N is the maximum allowed depth (root at depth 0 or 1?), and that deleting a node removes its entire subtree. Ask about input size and whether recursion depth is a concern.

2. Define the greedy strategy

Process nodes bottom-up (post-order). For each node, compute the height of its subtree. If the height exceeds N, delete the node (i.e., remove its subtree) and increment the deletion count; otherwise, keep it.

3. Prove optimality

Argue that deleting a node with height > N is always optimal: any valid solution must delete some node in that subtree, and deleting the highest such node removes more nodes and reduces depth at least as much. Use an exchange argument.

4. Analyze complexity

Each node is visited once, so time complexity is O(V) for V nodes. Space complexity is O(H) for recursion stack (H = tree height) or O(V) for an explicit stack/queue if iterative.

5. Discuss edge cases and implementation

Handle N=0 (delete all nodes except root? or delete root?), single-node tree, and deep trees causing stack overflow. Suggest iterative post-order traversal if recursion depth is a concern.

Key Points to Mention

Greedy bottom-up approach: delete nodes with subtree height > N.
Optimality proof via exchange argument: deleting a higher node is never worse.
Time complexity O(V), space complexity O(H) or O(V).
Edge cases: N=0, N >= tree height, single node, skewed tree.
Implementation details: post-order traversal, iterative vs recursive.
Connection to tree pruning and dynamic programming on trees.

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