The union-find angle came to me pretty fast, which was good.
Start by modeling the problem as a graph where customers are nodes and transactions are edges, then use a Union-Find (Disjoint Set Union) data structure to efficiently track connected components. Design a class that supports adding transactions (union operations) and querying connectivity (find operations) with near-constant time complexity. Discuss trade-offs between Union-Find and other approaches like BFS/DFS, and consider optimizations like path compression and union by rank.
Pro tip: Mention that Union-Find is ideal for dynamic connectivity but if the graph is static, precomputing components with BFS/DFS might be simpler; also highlight that path compression and union by rank make operations practically O(1).
Ask whether transactions are added incrementally or all at once, and whether queries are interleaved with updates. Clarify if customers are identified by IDs and if the graph is undirected (transactions imply mutual connection).
Select Union-Find (Disjoint Set Union) for dynamic connectivity due to its efficiency. Explain that each set represents a connected component, and union merges sets when a transaction occurs.
Define methods: addTransaction(customer1, customer2) to union two customers, and areConnected(customer1, customer2) to check if they share the same root. Include a constructor to initialize parent and rank arrays.
Use path compression in find to flatten the tree, and union by rank/size to keep trees shallow. This ensures nearly constant time per operation.
State that with optimizations, operations are O(α(n)) amortized, where α is the inverse Ackermann function. Compare with BFS/DFS which would be O(V+E) per query if graph is dynamic.
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Switched to BFS here and it felt like the right call.
Model the transaction network as a directed graph where customers are nodes and transactions are edges. Then perform a graph traversal (BFS or DFS) starting from the given customer, collecting all reachable nodes except the starting customer. Discuss handling cycles and large graphs efficiently.
Pro tip: Mention that you would use BFS with a visited set to avoid infinite loops in cyclic transaction networks, and consider distributed processing if the graph is too large for a single machine.
Confirm whether transactions are directed (e.g., payer to payee) and whether reachability should follow the direction of money flow. Ask about graph size and performance requirements.
Represent customers as nodes and transactions as directed edges. If transactions are bidirectional, treat edges as undirected.
Use BFS for shortest path or DFS for simplicity. Both work for reachability; BFS is often preferred for its iterative nature and ability to find shortest paths.
Start from the given customer, mark as visited, and explore neighbors. Add each newly visited node to the result set, excluding the start node.
Time complexity is O(V+E). Discuss handling cycles, disconnected components, and potential memory issues for large graphs.
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Start by clarifying the problem: the input is a graph of customers, a starting person, and a degree limit N. Then describe a BFS traversal that tracks depth and stops expanding beyond N, ensuring only nodes within N hops are returned. Discuss trade-offs like using BFS vs DFS, handling cycles, and potential optimizations for large graphs.
Pro tip: Mention that BFS naturally finds shortest paths in unweighted graphs, so it's ideal for degree-limited queries; also note that you can early-terminate when the queue's depth exceeds N to save work.
Confirm that the graph is unweighted, edges represent connections, and the degree limit N is inclusive (i.e., up to N hops). Ask about graph size, whether it's directed or undirected, and if the starting person is included in the result.
Explain that BFS explores nodes level by level, so the first time a node is visited, it's at its minimum hop distance from the start. This ensures we only include nodes within N hops.
Use a queue storing (node, depth) pairs, a visited set to avoid cycles, and a result list. Enqueue the start with depth 0, then while the queue is not empty, dequeue, add to result if depth ≤ N, and enqueue unvisited neighbors with depth+1 only if depth < N.
State that time complexity is O(V + E) in the worst case, but with degree limit N it's O(b^N) where b is branching factor, which can be much smaller. Space is O(V) for visited and queue. Mention that DFS could work but may not find shortest paths and could explore deeper unnecessarily.
For large graphs, consider bidirectional BFS if N is small, or using a depth-limited search with iterative deepening. Handle edge cases: N=0 (only start), disconnected graph, cycles, and self-loops.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.