Ziphq·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Jun 2026Remote

Summary

Ziphq had me do a tree DP problem that was basically House Robber III but extended to n-ary trees, which meant I also had to implement the tree structure from scratch. Not the hardest concept but the extra implementation overhead caught me a little off guard.

Questions Asked (1)

Given a general n-ary tree (which you must implement yourself, including the node structure with a children list), find the maximum amount you can 'rob' from nodes such that no two directly connected parent-child nodes are both robbed.

Algorithms & Data Structures

Author's notes

I knew House Robber III on a binary tree pretty well, so seeing it extended to n-ary felt manageable at first.

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AI HintsAI Generated

Suggested Approach

Model the problem as a tree DP where each node returns two values: the maximum sum when the node is robbed and when it is not. Use post-order traversal to combine children's results: if the node is robbed, children cannot be robbed; if not, take the max of each child's two states. Finally, return the max of the root's two states.

Pro tip: Explicitly define the two DP states and show how they transition; this demonstrates you understand the core recurrence and can avoid common pitfalls like double-counting or missing the 'skip' case.

1. Define the node structure

Implement a TreeNode class with a value and a list of children to represent the n-ary tree.

2. Define DP states

For each node, compute two values: max sum if the node is robbed (rob) and max sum if it is not robbed (skip).

3. Post-order traversal

Recursively process all children first, then compute the current node's rob and skip values using children's results.

4. Combine results

For rob: node value + sum of children's skip. For skip: sum of max(rob, skip) for each child.

5. Return final answer

After processing the root, return max(root.rob, root.skip).

Key Points to Mention

Dynamic programming on trees with two states per node
Post-order traversal to ensure children are processed before parent
Recurrence relation: rob = node.val + sum(child.skip), skip = sum(max(child.rob, child.skip))
Time complexity O(n) and space complexity O(h) for recursion stack
Handling edge cases: empty tree, single node, nodes with many children
Avoiding global state; returning a pair of values from each recursive call

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