I went with BFS flood-fill because I can write that cleanly under pressure and it passed.
Model the products as nodes in a graph where each pair represents an edge, then use Union-Find (Disjoint Set Union) to efficiently merge products into categories. After processing all pairs, count the distinct roots and their sizes to get the number of categories and their sizes.
Pro tip: Mention that Union-Find with path compression and union by rank achieves near O(1) amortized time per operation, making it optimal for large datasets. Also, clarify how you handle products that appear in no pairs—they form singleton categories.
Confirm that each pair indicates the two products belong to the same category, and that categories are transitive (if A shares with B and B with C, then A, B, and C are in the same category). Ask if products not appearing in any pair should be counted as individual categories.
Select Union-Find (Disjoint Set Union) to efficiently manage merging of categories. Explain that it supports union and find operations in nearly constant amortized time with path compression and union by rank.
Iterate through each product pair, perform union on the two products, and ensure all products are added to the Union-Find structure (including those not in any pair).
After processing, traverse all products, find their root, and tally the size of each root. The number of distinct roots is the number of categories, and the tally gives the size of each.
State that time complexity is O(N α(N)) where N is the number of products, and space is O(N). Discuss edge cases like empty input, duplicate pairs, and self-pairs.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Straightforward version of the same problem.
Start by clarifying the graph type (directed vs. undirected) and constraints, then propose either Union-Find or DFS/BFS as the core algorithm. Walk through the chosen approach step-by-step, analyze time and space complexity, and discuss edge cases like isolated nodes or empty graphs.
Pro tip: Mention that Union-Find with path compression and union by rank is often preferred for dynamic graphs or when edges are streamed, while DFS/BFS is simpler for static graphs. This shows you consider trade-offs beyond just correctness.
Confirm whether the graph is directed or undirected, if nodes are labeled 0 to n-1, and if there are any constraints on edge count or memory.
Select Union-Find (disjoint set) or graph traversal (DFS/BFS) based on graph properties and constraints. Explain why.
Describe the steps: initialize data structures, process edges or traverse, and count components. For Union-Find, detail union and find operations.
State time and space complexity. For Union-Find with optimizations, it's nearly O(E) time and O(n) space; for DFS/BFS, O(n+E) time and O(n) space.
Discuss cases like no edges (n components), complete graph (1 component), disconnected nodes, and potential recursion depth issues with DFS.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This was a follow-up and I fumbled for a second.
Clarify whether the graph is directed or undirected, as the cycle detection method differs. Then explain the appropriate algorithm: for undirected graphs, use Union-Find to check if the edge connects two vertices already in the same component; for directed graphs, use DFS to check if there's a path from the target to the source. Discuss time and space complexity and potential optimizations.
Pro tip: Mention that for undirected graphs, Union-Find with path compression and union by rank gives near-constant time per operation, making it ideal for dynamic edge additions. For directed graphs, note that DFS can be optimized with iterative implementation to avoid recursion limits.
Ask if the graph is directed or undirected, and whether it's static or dynamic. This determines the algorithm choice.
Maintain disjoint sets of vertices. For a new edge (u, v), check if u and v are in the same set; if yes, adding the edge creates a cycle. Otherwise, union the sets.
To check if adding edge (u, v) creates a cycle, check if there is already a path from v to u. If yes, adding (u, v) creates a cycle. Use DFS or BFS from v to see if u is reachable.
Discuss time and space complexity: Union-Find with path compression and union by rank is nearly O(1) per operation; DFS is O(V+E) per query. Mention trade-offs for dynamic graphs.
For multiple edge additions, consider incremental algorithms or maintaining a topological order for directed graphs. Handle self-loops and parallel edges.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.