The horse theme threw me off for a second but once I drew it out it's just a DAG reachability problem.
Model the relationships as a directed graph where edges point from parent to child, then determine if the two horses are connected in the underlying undirected graph. Use BFS/DFS from one horse to see if the other is reachable, or compute connected components to answer multiple queries efficiently.
Pro tip: Clarify whether the relationship is directed (ancestor/descendant) or undirected (shared ancestor) — the problem states both count, so treat it as undirected connectivity. Also discuss handling multiple queries by precomputing connected components or using union-find for near-constant time lookups.
Confirm that 'related' includes both direct ancestry and shared ancestors, meaning any path in the undirected graph. Ask about input size, number of queries, and whether the graph is a forest (each horse has at most one parent).
Represent the parent-child relationships as an adjacency list for a graph. If multiple queries, consider building a union-find structure or precomputing connected components.
For a single query, perform BFS/DFS from one horse to check if the other is reachable. For multiple queries, use union-find to group horses into connected components, then answer each query by checking if both horses belong to the same component.
For BFS/DFS: O(V+E) per query. For union-find: O(α(N)) per query after O(N) preprocessing. Discuss trade-offs and choose based on query frequency.
Consider cases where horses are the same, no relationship exists, or the graph has cycles (though typically it's a forest). Ensure the solution works for disconnected components.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.