Build referral chains and detect cycles

You are given a directed referral graph where each user may have at most one referrer. Input file 'referrals.csv' has columns: user_id (INT), referred_by (INT, nullable). Assumptions: up to 1e6 users; referred_by may reference a user_id not present in the file; rows may contain duplicates; self-referrals and multi-node cycles may exist. Write Python code to accomplish all of the following: 1) Implement chain(u) that returns the referral chain for user u from the earliest ancestor (root) to u as a list of user_ids. If any cycle is encountered on the path, detect it and return both: (a) the simple cycle nodes in encounter order, and (b) the acyclic prefix leading into the cycle; avoid infinite loops. 2) In O(n) time and O(n) memory overall, compute for every user: (root_ancestor[u], chain_depth[u]) where chain_depth[u] is the number of unique referrers on the path to the root. Do not recompute paths from scratch per user; use memoization/union-find/DFS with coloring (your choice) and justify complexity. 3) Return the top-3 longest valid acyclic chains as lists. Break ties by smaller root_ancestor id, then lexicographically smaller full chain list. 4) Robustness: describe (and implement) preprocessing to (a) deduplicate rows, keeping the earliest seen parent for a user if duplicates conflict, (b) normalize null/empty referred_by, (c) tolerate referred_by not found in user_id (treat as external root), and (d) flag self-referrals. 5) Provide minimal unit tests that cover: acyclic chain, self-cycle, and a 2-node cycle. Tiny example (use exactly these rows for tests): referrals.csv user_id,referred_by 1, 2,1 3,1 4,2 5,4 6,5 7,7 8,9 9,8 Expected behaviors to assert (do not print answers here): - chain(6) traverses 1→2→4→5→6; depth(6)=4; root=1 - chain(7) reports a self-cycle at 7 - chain(8) reports the 2-node cycle {8,9}.