bfs

Graph traversal sits beneath a large family of interview problems. What matters is knowing the BFS-vs-DFS tradeoff (shortest path vs memory shape) and applying it cleanly, then getting the directed-vs-undirected cycle gotcha that trips up most candidates.

A staple grid-traversal problem probing BFS/DFS flood fill and disciplined visited-marking. What interviewers watch for: viewing the grid as a graph and tallying connected components. Here is the answer.

Turning a tree into a string and reconstructing it faithfully rests on one choice most candidates overlook. Skip it and the structure turns ambiguous. Here is the answer and the interview-grade reasoning.

The words encode a partial order among characters. Convert each adjacent pair into a directed edge, then topologically sort. The traps are the prefix edge case and spotting contradictions. Here is the full solution with cycle handling.

Treat each word as a node and one-letter edits as edges: the shortest ladder becomes a shortest path in an unweighted graph, which means BFS. The detail that decides pass or fail is producing neighbors in O(26 * L) without scanning the entire dictionary. Here is the pattern.