Pretty approachable as a standalone problem.
Clarify the input format and constraints, then propose an efficient algorithm that computes weighted match scores between the target user and all other records, filtering those that meet the threshold. Discuss trade-offs between time and space complexity, and consider optimizations like indexing or early termination.
Pro tip: Mention that in a real system like Stripe, you'd likely precompute and index similarity scores or use a inverted index for scalability, showing awareness of production concerns beyond the basic algorithm.
Ask about input size, field types, weight normalization, threshold range, and whether records can have missing fields. Confirm output format (list of IDs).
Explain how to compute the weighted score: for each field, compare values (e.g., exact match, case-insensitive, or fuzzy) and multiply by weight, then sum. Normalize if needed.
Propose a straightforward O(n) approach iterating over all records, computing scores, and collecting IDs above threshold. Discuss potential optimizations for large n.
State time and space complexity. Discuss alternatives like indexing, blocking, or approximate matching if scale is large, and trade-offs between accuracy and performance.
Consider missing fields, zero weights, threshold boundaries, and duplicate records. Outline test cases to validate correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This is where I started second-guessing my data structures.
Use a breadth-first search (BFS) from the target user, expanding exactly two levels while tracking visited nodes to avoid duplicates. At each level, collect all neighbors of the current frontier, ensuring that only nodes at distance 2 are added to the result set. Return the union of direct matches (distance 1) and second-hop matches (distance 2).
Pro tip: Clarify whether the result should include the target user or exclude it, and discuss how to handle cycles or duplicate edges to show attention to edge cases. Also, mention that for large graphs, a distributed BFS or precomputed adjacency lists can improve performance.
Confirm the definition of 'exactly two hops', whether the target user is included, and how to handle duplicate records or cycles. Ask about graph size and performance constraints.
Select BFS because it naturally explores level by level, making it easy to stop at depth 2. Explain why DFS would be less efficient for this exact-hop requirement.
Initialize a queue with the target user at depth 0. Process nodes level by level, incrementing depth after each level. Stop when depth reaches 2, collecting all nodes at depth 1 and 2.
Use a visited set to avoid revisiting nodes and to prevent infinite loops in cyclic graphs. Ensure the result set contains unique records and excludes the target user if required.
Discuss time and space complexity: O(V + E) for BFS, where V is vertices and E is edges. Mention potential optimizations like bidirectional BFS or precomputed adjacency lists for large-scale systems.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Full BFS or DFS from the target, return everything visited except the target itself.
Clarify the graph representation and constraints, then choose between BFS/DFS for traversal. Discuss trade-offs like recursion depth, memory usage, and whether to use iterative BFS to avoid stack overflow. Finally, outline the algorithm and analyze time/space complexity.
Pro tip: Mention that BFS is generally preferred for finding all reachable nodes because it avoids recursion limits and can be more cache-friendly. Also, highlight the importance of marking nodes as visited to prevent infinite loops in cyclic graphs.
Ask about graph size, whether it's directed or undirected, and if there are cycles. Confirm that we need all nodes in the connected component, not just the shortest path.
Decide between BFS and DFS based on constraints. BFS is iterative and avoids recursion depth issues; DFS is simpler but may overflow stack for large graphs.
Initialize a queue (for BFS) or stack (for DFS) with the target user, and a visited set. While the queue is not empty, pop a node, add it to the result, and enqueue all unvisited neighbors.
Time complexity is O(V + E) where V is vertices and E is edges. Space complexity is O(V) for visited set and queue/stack. Discuss trade-offs between BFS and DFS in terms of memory and recursion.
Consider disconnected graphs, self-loops, and large graphs. Mention possible optimizations like early termination if only a subset is needed, or using union-find for multiple queries.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.