Databricks·Software Engineer·Technical Phone Screen·Senior

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Jun 2026Remote

Summary

Databricks threw a tree problem at me that looked like a standard traversal question until I read the constraint about not being allowed to define a node class. The whole thing had to be done with raw integer arithmetic on preorder ids, which is a very different beast.

Questions Asked (1)

Given a k-order Fibonacci tree where nodes are labeled 0 through size-1 in preorder, find the shortest path between two nodes by their ids. You cannot define a node class and must work purely with integer arithmetic based on the tree's structural properties.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

I stared at this for a solid minute because my instinct was to build a tree and do parent pointers.

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AI HintsAI Generated

Suggested Approach

First, derive the recursive size formula for a k-order Fibonacci tree and use it to map any preorder label to its position in the tree. Then, compute the path between two nodes by repeatedly moving the deeper node to its parent using integer arithmetic, leveraging the tree's self-similar structure. Finally, count the steps to find the shortest path length.

Pro tip: Emphasize that the preorder labeling allows you to determine a node's subtree boundaries and parent without explicit pointers, which is crucial for the integer-only constraint. Also, mention that the shortest path in a tree is simply the unique path, so the problem reduces to finding the lowest common ancestor efficiently.

1. Understand the k-order Fibonacci tree structure

Define the tree recursively: a k-order Fibonacci tree of order n has a root with k subtrees, each being a k-order Fibonacci tree of orders n-1, n-2, ..., n-k. Derive the size formula S(n) = 1 + S(n-1) + S(n-2) + ... + S(n-k) with base cases.

2. Map preorder label to node position

Given a node id (preorder index), determine its depth and which subtree it belongs to by comparing the id against cumulative subtree sizes. This allows you to find its parent and children using integer arithmetic.

3. Compute the path between two nodes

While the two nodes are not equal, move the one with greater depth to its parent. If depths are equal, move both to their parents. Count each move as one edge in the path.

4. Optimize using LCA

Instead of moving step by step, find the lowest common ancestor (LCA) by aligning depths and then moving both nodes up simultaneously until they meet. The path length is depth(a) + depth(b) - 2*depth(LCA).

5. Handle edge cases and validate

Consider cases where one node is an ancestor of the other, or when nodes are the same. Validate with small k and n values by manually constructing the tree.

Key Points to Mention

Recursive size formula for k-order Fibonacci tree and its closed form or efficient computation.
Preorder labeling property: each subtree occupies a contiguous range of labels.
Finding parent and children using integer arithmetic and subtree sizes.
Shortest path in a tree is the unique path, so LCA is key.
Time complexity: O(depth) per query, which is O(n) in worst case, but can be optimized with binary lifting if needed.
Space complexity: O(1) extra space if computing on the fly, or O(n) for precomputed sizes.

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