Google·Software Engineer·Onsite - Coding / Algorithms·Intermediate

IntermediatePrefer not to say

Apr 2026

Summary

Google SWE coding round, one problem the whole session. Graph/clustering problem that looked straightforward but has some edge cases that'll bite you if you're not careful about how you handle transitivity.

Questions Asked (1)

Given a list of 2D points and a radius r, group the points into clusters where two points belong to the same cluster if the Euclidean distance between them is at most r, and cluster membership is transitive. Return the total number of distinct clusters.

Algorithms & Data Structures

Author's notes

My first instinct was BFS and it was the right call, but I wasted probably five minutes second-guessing myself and almost went down a sorting rabbit hole.

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AI HintsAI Generated

Suggested Approach

Model the points as a graph where edges connect points within distance r, then count connected components using Union-Find (DSU) or BFS/DFS. For efficiency, avoid O(n^2) pairwise checks by using a spatial index like a grid or k-d tree to find neighbors within radius r.

Pro tip: Mention the trade-offs between Union-Find and BFS/DFS, and highlight that using a spatial grid can reduce time complexity from O(n^2) to O(n) in practice for uniformly distributed points. Also, clarify that the problem is equivalent to finding connected components in a unit disk graph.

1. Clarify and Restate

Confirm the problem: points are in 2D, distance threshold r, transitive clustering. Ask about input size, coordinate ranges, and whether points can be duplicated.

2. Choose Data Structure

Decide between Union-Find (DSU) for dynamic connectivity or BFS/DFS for static graph. Consider using a spatial index (grid, k-d tree) to efficiently find neighbors within r.

3. Build Graph or Process Neighbors

For each point, find all other points within distance r using the spatial index. Union them if using DSU, or add edges if building a graph.

4. Count Clusters

If using DSU, count the number of distinct roots. If using BFS/DFS, count the number of times you initiate a traversal from an unvisited point.

5. Analyze Complexity and Optimize

Discuss time and space complexity. Naive pairwise is O(n^2); with spatial partitioning, it can be near O(n) for uniform points. Mention worst-case scenarios.

Key Points to Mention

Union-Find (Disjoint Set Union) with path compression and union by rank for efficient connectivity queries.
Spatial indexing techniques: uniform grid, k-d tree, or sweep line to avoid O(n^2) neighbor checks.
Graph traversal (BFS/DFS) as an alternative to Union-Find, especially when the graph is sparse.
Time complexity analysis: O(n^2) naive vs. O(n log n) or O(n) with spatial partitioning.
Edge cases: no points, all points within r, points exactly at distance r, duplicate points.
Transitivity of clustering: connected components in the graph formed by edges between points within distance r.

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