BFS was the obvious call here and I got there pretty fast.
Model the grid as a graph and use BFS from the starting position to find the nearest meeting room, as BFS guarantees the shortest path in an unweighted grid. Early exit when the first meeting room is dequeued, and return [-1, -1] if the queue empties without finding one.
Pro tip: Mention that BFS is optimal here because all moves have equal cost, and note that early termination can significantly reduce runtime in practice. Also, clarify that you handle edge cases like the start being a meeting room or an invalid position.
Confirm the grid dimensions, movement directions, and what constitutes a valid move. Ask about edge cases such as starting on a wall or meeting room, and whether the grid can be modified.
Explain that BFS is ideal for finding the shortest path in an unweighted grid. Justify why DFS or Dijkstra would be less efficient or unnecessary.
Describe initializing a queue with the start position, a visited set to avoid revisiting cells, and processing neighbors in four directions. Check for meeting rooms upon dequeue or enqueue.
State that time complexity is O(m*n) since each cell is visited at most once, and space complexity is O(m*n) for the queue and visited set in the worst case.
Mention early termination when a meeting room is found, and handling cases where no meeting room is reachable. Optionally, discuss bidirectional BFS or multi-source BFS if multiple starts are considered.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Model the environment as a graph where rooms are nodes and connections are edges, then run BFS from each user to compute shortest distances to all reachable rooms. Iterate over the intersection of reachable rooms to find the one minimizing the sum of distances, returning its coordinates and the total distance.
Pro tip: Clarify upfront whether the graph is unweighted (BFS) or weighted (Dijkstra), and mention that if the graph is large but queries are frequent, precomputing all-pairs shortest paths or using landmarks can be beneficial.
Confirm whether the graph is unweighted or weighted, directed or undirected, and how rooms/coordinates are represented. Ask about constraints (e.g., number of rooms, edges) to choose the right algorithm.
Run BFS (for unweighted) or Dijkstra (for weighted) from each user's starting position to compute distances to all reachable rooms. Store distances in hash maps keyed by room ID.
Take the intersection of the two distance maps to get rooms reachable by both users. If the intersection is empty, return no meeting room.
Iterate over the common rooms, compute the sum of distances for each, and track the room with the minimum total. Return its coordinates and the minimum sum.
State time complexity O(V+E) for BFS per user, plus O(V) for intersection and minimization. Mention potential optimizations like bidirectional BFS or precomputation for multiple queries.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.