Atlassian·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Apr 2026

Summary

Atlassian SWE interview that came down to a tree problem, specifically finding the lowest common ancestor between two leaf nodes in a company org chart modeled as a tree. Pretty focused session, one meaty algorithmic question with a complexity discussion tacked on.

Questions Asked (1)

You're given a tree where internal nodes are departments and leaves are employees. Given two employee leaf nodes, find the deepest department that contains both of them. Also walk through the time and space complexity of your solution.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

Classic LCA but dressed up in org chart clothing, which threw me for a second because I was thinking about it in terms of HR logic before I realized it's just lowest common ancestor on a tree.

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AI HintsAI Generated

Suggested Approach

Model the tree and find the lowest common ancestor (LCA) of the two employee nodes, which is the deepest department containing both. Use an efficient LCA algorithm like binary lifting or Euler tour + RMQ, and analyze time and space complexity.

Pro tip: Clarify whether the tree is static or dynamic; if static, preprocess for O(1) LCA queries, but if dynamic, consider a different approach. Also, mention that the LCA is the deepest node that is an ancestor of both leaves.

1. Understand the problem

Restate the problem: given a tree with internal nodes as departments and leaves as employees, find the deepest department (internal node) that is an ancestor of both given employee leaves. This is equivalent to finding the lowest common ancestor (LCA) of the two leaves.

2. Choose an LCA algorithm

Select an appropriate LCA algorithm based on constraints. For static trees, binary lifting (O(n log n) preprocessing, O(log n) query) or Euler tour + RMQ (O(n) preprocessing, O(1) query) are common. For dynamic trees, consider other methods.

3. Implement the solution

Implement the chosen algorithm: preprocess the tree to answer LCA queries, then for the given two leaves, compute their LCA. Ensure the LCA is an internal node (department) and not a leaf.

4. Analyze complexity

Clearly state the time and space complexity of preprocessing and querying. For binary lifting: O(n log n) time and space for preprocessing, O(log n) per query. For Euler tour + RMQ: O(n) preprocessing time and space, O(1) per query.

5. Discuss trade-offs

Compare approaches: binary lifting is simpler to implement and works for dynamic trees if updates are infrequent; Euler tour + RMQ gives faster queries but requires more complex preprocessing and is less flexible for updates. Choose based on expected query frequency and tree dynamics.

Key Points to Mention

Lowest Common Ancestor (LCA) concept and its equivalence to the deepest department containing both employees.
Binary lifting technique: preprocessing with ancestors at powers of two, and querying by lifting nodes to the same depth and then together.
Euler tour + RMQ approach: flatten tree into an array and use segment tree/sparse table for range minimum queries on depths.
Time and space complexity analysis for each approach, including preprocessing and per-query costs.
Handling edge cases: when one employee is an ancestor of the other (though leaves cannot be ancestors, but in general), or when the tree is skewed.
Trade-offs between preprocessing time, query time, and space, and considerations for dynamic updates.

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