Reinforce Labs·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Interviewed for a software engineer role at Reinforce Labs and got hit with a tree problem that looked straightforward but had a real gotcha in the approach. The algorithmic depth they expected was higher than I anticipated.

Questions Asked (1)

Given a rooted tree where each node holds a count of stones, make the tree 'balanced' by ensuring every internal node's direct children subtrees all have equal total stone counts. You can only add stones (each addition costs 1). Return the minimum number of stones to add.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

My first instinct was top-down DFS and I spent a few minutes going down that path before realizing it doesn't work.

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AI HintsAI Generated

Suggested Approach

Use a post-order DFS to compute the total stones in each subtree. For each internal node, determine the maximum subtree sum among its children, then add stones to the other child subtrees to match that maximum. Accumulate the total additions and return the sum.

Pro tip: Clarify that the tree is rooted and that 'balanced' applies only to internal nodes, not leaves. Also, mention that the greedy choice of raising all children to the maximum is optimal because adding stones to the largest subtree would only increase the required additions elsewhere.

1. Understand the problem and constraints

Confirm that only additions are allowed, each costing 1, and that the goal is to minimize total additions. Ensure you know the tree structure and that leaves are exempt from balancing.

2. Design a recursive post-order traversal

Traverse the tree bottom-up, computing for each node the total stones in its subtree after balancing. For leaves, the total is just the node's stone count.

3. Balance each internal node

For an internal node, collect the balanced subtree sums of its children. Find the maximum among them, then add stones to each other child subtree to reach that maximum. Sum these additions.

4. Propagate and accumulate

Return the new total stones for the subtree (node's own stones plus the sum of balanced children sums) and accumulate the total additions across all internal nodes.

5. Analyze complexity and edge cases

State that the algorithm runs in O(n) time and O(h) space for recursion. Discuss edge cases like single-node trees, skewed trees, and large stone counts.

Key Points to Mention

Post-order DFS to compute subtree sums bottom-up.
Greedy strategy: raise all children to the maximum child subtree sum.
Proof of optimality: adding to the maximum would require even more additions elsewhere.
Time complexity O(n) and space complexity O(h) due to recursion stack.
Handling of leaf nodes (no balancing needed).
Potential integer overflow and use of 64-bit integers for large sums.

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