Classic BFS problem once you see it, but I didn't see it right away.
Clarify the lock's mechanics and the definition of a 'turn' (e.g., rotating one dial by one position). Model the problem as a shortest path on a state graph where nodes are lock combinations and edges are single turns, then apply BFS to find the minimum number of turns from the initial to the target combination.
Pro tip: Demonstrate maturity by proactively discussing constraints and edge cases (e.g., deadends, circular dials, multiple dials) and comparing BFS with bidirectional BFS or A* for efficiency.
Ask questions to understand the lock's structure (number of dials, positions per dial), what constitutes a turn, and whether there are forbidden combinations. Confirm the goal is to minimize the number of turns.
Represent each possible combination as a node. Connect two nodes with an edge if one can be reached from the other by a single turn (e.g., rotating one dial by one position).
Since each turn has equal cost, BFS from the initial combination will find the minimum number of turns to reach the target. Use a queue and a visited set to avoid cycles.
For large state spaces, consider bidirectional BFS (search from both start and target) or A* with a heuristic like the sum of minimal dial rotations. Discuss trade-offs.
State time and space complexity (O(N) where N is number of reachable states). Handle edge cases: start equals target, deadends, circular dials (modulo arithmetic), and multiple dials.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.