I jumped straight to BFS which is right, but my first state representation was just (row, col) and the interviewer let me run with it for a bit before asking what happens when you visit the same cell twice but with different keys collected.
Model the problem as a shortest path search in a state space where each state is (row, col, keys_collected_bitmask). Use BFS to find the minimum steps to reach any state with all keys collected. Explain how you handle doors, keys, and revisiting states efficiently.
Pro tip: Emphasize that BFS is optimal for unweighted grids and that the bitmask state space is bounded by m*n*2^6, making it efficient. Mention that you can prune by not revisiting the same (position, keys) state, and that you can precompute key and door locations to speed up checks.
Represent each state as (row, col, keys_bitmask) where keys_bitmask is a 6-bit integer indicating which keys (a-f) have been collected. This captures all necessary information to determine future moves.
Use BFS because each move costs 1 step and we need the minimum steps. BFS explores states in increasing order of steps, guaranteeing the first time we reach a state with all keys is optimal.
From a state, try moving in four directions. If the cell is a wall, skip. If it's a door (A-F), only proceed if the corresponding key is in the bitmask. If it's a key (a-f), update the bitmask by setting the corresponding bit.
Maintain a visited set or 3D boolean array of size m×n×64 to avoid revisiting the same (row, col, keys) state. This prevents cycles and ensures efficiency. Optionally, prune states that cannot possibly collect all keys (e.g., if some keys are unreachable).
Time complexity: O(m*n*2^K) where K is number of keys (≤6). Space complexity: O(m*n*2^K) for the visited structure and queue. This is efficient because 2^6=64, so at most 64*m*n states.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.