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My first instinct was to compute depths and find the lowest common ancestor, which works but I fumbled explaining the LCA part clearly under pressure.
Model the hierarchy as an undirected tree, then find the shortest path between the two employees. Use LCA (Lowest Common Ancestor) to compute the distance as depth(a) + depth(b) - 2*depth(lca). Alternatively, BFS from one employee to the other works but is less efficient for repeated queries.
Pro tip: Clarify whether the hierarchy is guaranteed to be a tree (single root, no cycles) and whether the two employees are always in the same tree. Mention that if the tree is large and queries are frequent, precomputing depths and binary lifting for LCA gives O(log n) per query.
Confirm that the pairs form a valid tree (one root, no cycles) and that both employees exist. Ask if the tree is static or dynamic, and if multiple queries will be made.
Construct an adjacency list from the manager-employee pairs, treating edges as undirected. Identify the root (the employee with no manager) if needed.
For a single query, BFS from one employee to the other is simple and O(n). For multiple queries, preprocess for LCA (e.g., binary lifting) to answer in O(log n) per query.
If using LCA, compute depths of both nodes and their LCA, then distance = depth(a) + depth(b) - 2*depth(lca). If using BFS, track distance during traversal.
Discuss time/space complexity. Handle edge cases: same employee (distance 0), one is ancestor of the other, or disconnected (if not guaranteed tree).
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The set-based version was easy, just track visited numbers and bail if you see a repeat that isn't 1.
Start by clarifying the problem and edge cases, then present the two required solutions: one using a hash set to detect cycles, and another using Floyd's cycle detection algorithm for O(1) space. Compare their time and space complexities, and discuss when each is preferable.
Pro tip: Mention that the sum of squares operation always leads to a cycle, and the only cycle that reaches 1 is the happy number cycle; all other cycles never include 1. This shows deep understanding and can help justify the cycle detection approach.
Restate the problem: given a positive integer, repeatedly replace it with the sum of the squares of its digits, and determine if it eventually becomes 1. Ask about constraints (e.g., input size, negative numbers) and confirm that the process always enters a cycle.
Use a hash set to store numbers encountered. At each step, compute the sum of squares of digits; if it's 1, return true; if it's already in the set, return false; otherwise add it and continue. Analyze time complexity: O(log n) per step, but number of steps is bounded by a constant for 32-bit integers.
Apply Floyd's tortoise and hare algorithm: use two pointers, slow and fast, where slow advances one step and fast advances two steps. If they meet, a cycle exists; if fast reaches 1, return true. This uses constant extra space.
Compare the two solutions: the set approach is simpler but uses O(k) space where k is the cycle length or path length; Floyd's uses O(1) space but may do more computations. Discuss edge cases: input 1 (immediately happy), input 0 (not positive, but if allowed, 0 leads to 0 cycle), and large numbers.
Summarize that both solutions are valid; the set approach is easier to implement and understand, while Floyd's is more space-efficient. In an interview, mention that for 32-bit integers, the maximum sum of squares is bounded (e.g., 9^2 * 10 = 810), so the set size is small, making the set approach practically O(1) space as well.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.