The example they gave was small enough that I almost just hardcoded a BFS and called it a day.
Model the problem as finding a meeting node M that minimizes dist(A, M) + dist(B, M) + 2 * dist(M, D), where dist is the shortest path distance. Precompute shortest paths from A, B, and D using BFS (for unweighted) or Dijkstra (for weighted), then iterate over all nodes to find the minimum. Also consider the case where they never meet (M = D) as a special case.
Pro tip: Clarify with the interviewer whether the graph is weighted or unweighted, as this determines the algorithm choice. Mentioning edge cases like disconnected graphs or when A or B is already at D shows thoroughness.
Ask about graph size, weighted vs unweighted edges, and whether nodes can be revisited. Confirm that the goal is to minimize the sum of distances traveled by both agents.
Let M be the node where they meet (M could be D if they never meet). The total distance is dist(A, M) + dist(B, M) + 2 * dist(M, D). Explain why the factor 2 appears: both travel from M to D together.
Run BFS from A, B, and D (or Dijkstra if weighted) to get distances from each to all nodes. This takes O(V+E) per BFS for unweighted graphs.
For each node M, compute the total distance using the precomputed distances, and track the minimum. Also consider M = D as the no-meet case.
State time complexity O(V+E) for unweighted (or O((V+E) log V) for weighted) and space O(V). Discuss handling disconnected graphs (return infinity) and nodes already at D.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.