I went in thinking BFS, which was right, but I underestimated how much state you need to track.
Model the problem as a graph traversal where crates are nodes and keys/accessibility define edges. Use a worklist algorithm with separate queues for unlocked and locked crates, processing unlocked crates immediately and deferring locked ones until their key is found. Track visited crates to avoid cycles and duplicates, and accumulate tokens as you open crates.
Pro tip: Emphasize that the algorithm must handle dynamic discovery: new crates and keys are added during traversal, so you need to check deferred locked crates whenever a new key is acquired. This shows you understand the incremental nature of the problem.
Represent each crate as a node with attributes: locked/unlocked status, token count, keys it contains, and crates it unlocks. Edges represent accessibility: from a crate to the crates it makes accessible, and from keys to the crates they unlock.
Use a queue for accessible unlocked crates, a map from key to locked crates waiting for that key, a set for visited crates, and a set for collected keys. Start by enqueuing all initially accessible unlocked crates.
While the queue is not empty, dequeue a crate, mark it visited, add its tokens, collect its keys, and for each key, unlock any waiting crates and enqueue them if they become accessible. Also enqueue any newly discovered crates that are unlocked and accessible.
When encountering a locked crate, if you have the key, unlock and enqueue it; otherwise, add it to a waiting list for that key. When a new key is found, check the waiting list and enqueue any crates that can now be unlocked.
Continue until the queue is empty. The total tokens collected is the sum of tokens from all visited crates. Ensure all crates are processed at most once to achieve O(n + total_edges) time.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.