The chicken framing is funny but don't let it distract you, it's just an ancestor graph problem.
Clarify that the relationships form a forest (or tree) and choose an appropriate data structure like adjacency lists for parent-to-child links. For the ancestor check, use DFS/BFS from one chicken to find all ancestors, then check if the other chicken is in that set; alternatively, find ancestors of both and intersect. Discuss time and space complexity based on the chosen approach.
Pro tip: Mention that if the tree is large and queries are frequent, you can preprocess with binary lifting or Euler tour + RMQ to answer ancestor queries in O(1) or O(log n) time. Also, consider edge cases like the same chicken, direct parent-child, and disconnected components.
Ask if the relationships form a tree or forest, if there are cycles, and if the chickens are identified by unique IDs. Confirm whether we need to handle multiple queries or just one.
Represent the relationships using an adjacency list (e.g., dictionary mapping parent to list of children) or a parent pointer map. For efficient ancestor checks, consider building a set of ancestors for one chicken.
Use DFS or BFS to traverse from one chicken up to the root, collecting all ancestors. Then check if the other chicken is in that set. Alternatively, traverse both upward simultaneously and look for intersection.
Time complexity: O(N) for traversal in the worst case, where N is the number of chickens. Space complexity: O(N) for storing ancestors or recursion stack. Mention that with preprocessing, queries can be faster.
For multiple queries, preprocess the tree with binary lifting or Euler tour to answer in O(log N) or O(1). Discuss trade-offs between preprocessing time and query time.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.