The chicken theme threw me for a second and I almost laughed, but it's really just 'are these two nodes connected in a directed graph.' I went with BFS from both nodes and checked for overlap in their ancestor/descendant sets.
Model the relationships as a graph and determine if two nodes are connected. Use union-find for efficient connectivity queries or BFS/DFS to traverse from one chicken to the other. Clarify assumptions about the dictionary structure and whether the graph is a tree or a general graph.
Pro tip: Discuss trade-offs between union-find and traversal: union-find is ideal for multiple queries, while BFS/DFS is simpler for a single query. Mention that if the graph is a tree, you can find the root and compare paths, but union-find handles general graphs.
Ask about the dictionary format (e.g., parent -> list of children), whether the graph is directed or undirected, if it's a tree or a general graph, and if there are multiple queries.
Decide between union-find (for multiple queries) and BFS/DFS (for a single query). Consider building an undirected graph if relationships are bidirectional.
For union-find, initialize each chicken as its own set, union all parent-child pairs, then check if the two chickens have the same root. For BFS/DFS, build an adjacency list and search from one chicken to the other.
Union-find with path compression and union by rank: O(α(N)) per operation, nearly constant. BFS/DFS: O(V+E) time and space. Discuss trade-offs.
Walk through a simple example, including edge cases like same chicken, disconnected components, and cycles if applicable.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.