The first two parts were fine, BFS is BFS.
Start by clarifying the problem constraints and then propose a unified BFS solution that naturally answers all four sub-questions: BFS gives reachability, shortest distance, and a shortest path via parent pointers; augment BFS to count shortest paths by summing counts from predecessors. Discuss trade-offs between BFS and other approaches (e.g., A* for pathfinding, DP for counting) and mention optimizations like bidirectional BFS for large grids.
Pro tip: Mention that the counting sub-question requires careful handling of multiple predecessors and that using a queue with distance tracking avoids revisiting nodes, ensuring O(mn) time. Also, note that returning a path can be done by backtracking from the target using parent pointers, and that storing parents for all nodes is O(mn) space, which is acceptable but can be optimized if only one path is needed.
Ask about grid size, movement directions (4 or 8), whether start/target can be blocked, and if the grid can be modified. Confirm that all sub-questions should be answered in sequence and that a single traversal is preferred.
Explain that BFS from the start explores cells in increasing distance order, so it can determine reachability, shortest distance, and a shortest path. For counting, augment BFS to track the number of shortest paths to each cell.
Use a queue for BFS, a distance array initialized to -1, and a parent array to reconstruct the path. When visiting a neighbor, if unvisited, set distance and parent; if already visited with distance+1, update path count by adding the current cell's count.
After BFS, check if target is reachable (distance != -1), report distance, reconstruct path by backtracking from target to start using parents, and report the path count stored at the target.
Mention that BFS is O(mn) time and space, which is optimal for unweighted grids. For very large grids, bidirectional BFS can reduce search space. For counting, if the grid is huge, consider modular arithmetic to avoid overflow.
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