Others·Data Scientist·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Took a technical screen for a Data Scientist role and got a tree traversal problem that looked deceptively clean on the surface. The core challenge was making parent-ancestor lookups fast enough for large inputs, which pushed me toward thinking about binary lifting rather than naive traversal.

Questions Asked (1)

Given a rooted tree represented as a parent array, answer multiple queries where each query asks: starting from node u, if you move up to the parent k times (stopping at root if you run out of ancestors), which node do you land on? Design an efficient solution for large n and q.

Algorithms & Data Structures

Author's notes

My first instinct was just to walk up the tree k steps per query, which is fine for small inputs but obviously falls apart at 2e5 nodes and queries.

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AI HintsAI Generated

Suggested Approach

Preprocess the tree to answer each query in O(log n) or O(1) time using binary lifting (upward table) or Euler tour + depth arrays. For each query, lift u by k steps using the precomputed ancestors, handling the case where k exceeds depth by returning the root. Discuss trade-offs between preprocessing time and query time.

Pro tip: Mention that binary lifting is a standard technique for ancestor queries and that it can be extended to support other operations like LCA. Also, note that if queries are offline, you can process them in O(n + q) using a DFS with a stack, which might be simpler and more efficient for certain constraints.

1. Clarify the problem and constraints

Restate the problem: given a parent array representing a rooted tree, answer queries (u, k) asking for the k-th ancestor of u. Ask about constraints on n and q to determine the required efficiency.

2. Choose a preprocessing strategy

Decide between binary lifting (O(n log n) preprocessing, O(log n) per query) or Euler tour + depth (O(n) preprocessing, O(1) per query with level ancestor data structure). For simplicity, binary lifting is often preferred.

3. Implement binary lifting

Build a table up[j][v] where up[0][v] is the parent of v, and up[j][v] = up[j-1][ up[j-1][v] ]. This allows jumping 2^j steps at once.

4. Answer queries efficiently

For each query (u, k), if k > depth[u], return root. Otherwise, iterate over bits of k from highest to lowest, and if the j-th bit is set, set u = up[j][u]. Finally, return u.

5. Analyze complexity and edge cases

Time: O(n log n) preprocessing, O(log n) per query. Space: O(n log n). Handle edge cases: k=0 (return u), u=root, and k larger than depth.

Key Points to Mention

Binary lifting (also known as jump pointers) for ancestor queries.
Time and space complexity: O(n log n) preprocessing, O(log n) per query, O(n log n) space.
Handling k larger than depth by returning the root.
Alternative approach: Euler tour + depth array with level ancestor data structure for O(1) queries.
Offline processing with DFS and stack for O(n + q) if queries are known in advance.
Edge cases: k=0, u=root, and multiple queries with same u.

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