My first instinct was BFS and it works fine, but I think they were nudging toward union-find.
Model the problem as finding connected components in an undirected graph where each person is a node and friendships are edges. Use either DFS/BFS to traverse each component or Union-Find to merge connected people, counting the number of components. Discuss trade-offs between the two approaches based on graph density and constraints.
Pro tip: Mention that the adjacency matrix is symmetric and represents an undirected graph, so you can optimize by only traversing the upper triangle or using Union-Find with path compression and union by rank for near-linear time. Also, clarify that the matrix is 1-indexed or 0-indexed and handle edge cases like n=0.
Confirm that the matrix represents an undirected graph, friendship is transitive, and ask about input size to choose the right algorithm. Clarify if the matrix is guaranteed symmetric and if self-friendship (diagonal) is always 1.
Decide between DFS/BFS (simpler, O(n^2) time, O(n) space) and Union-Find (efficient for dynamic connectivity, O(n^2 α(n)) time, O(n) space). Explain your choice based on constraints.
For DFS/BFS: iterate through each person, if unvisited, increment circle count and traverse all connected friends. For Union-Find: initialize each person as its own parent, union all friends, then count unique roots.
State time and space complexity. Discuss edge cases: n=0, n=1, all friends (one circle), no friends (n circles), and disconnected groups.
Walk through a small example (e.g., 3x3 matrix) to verify correctness. If time permits, mention potential optimizations like early termination or using bitsets for dense graphs.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.