← Pinterest Interview Insights
I stared at this for a bit before realizing it's a graph reachability problem.
Model the problem as a graph where pins are nodes and boards are hyperedges connecting pins. Use union-find to efficiently determine connectivity, and for the follow-up, compute the shortest path in the bipartite graph of pins and boards using BFS.
Pro tip: Clarify whether the graph is static or dynamic; if boards are frequently updated, consider incremental union-find or caching connectivity results. Also, mention that the minimum number of boards is equivalent to the shortest path length in the bipartite graph minus one, divided by two.
Ask about the size of the data, whether the graph is static or dynamic, and if the function needs to handle multiple queries. Confirm that 'connected' means there exists a sequence of pins and boards linking the two pins.
Represent pins as nodes and boards as hyperedges that connect all pins on that board. Alternatively, create a bipartite graph with pins and boards as nodes, and edges between a pin and a board if the pin belongs to that board.
Use union-find (disjoint set union) to group pins that are connected. Iterate through each board, union all pins on that board. Then check if the two pins have the same root.
For the minimum number of boards, perform BFS on the bipartite graph from one pin to the other. The number of boards in the path is (path length - 1) / 2, since the path alternates between pins and boards.
Discuss time and space complexity: union-find is near O(N α(N)) for N pins, BFS is O(V+E). For multiple queries, precompute connected components and store distances or use bidirectional BFS for efficiency.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.