I knew this was a graph problem pretty quickly but fumbled explaining why BFS and not DFS.
Model the problem as a graph where each word is a node and edges connect words differing by one letter. Use BFS from the start word to find the shortest path to the end word, counting levels as transformation steps. Return 0 if the end word is never reached.
Pro tip: Mention bidirectional BFS as an optimization: it significantly reduces the search space by expanding from both ends and meeting in the middle, which is especially important for large dictionaries. Also, clarify edge cases like start == end (return 1) and words not in dictionary.
Confirm constraints: dictionary size, word length, case sensitivity, and whether start/end must be in dictionary. Handle edge cases: start equals end, end not in dictionary, or no possible path.
Explain that each word is a node and edges exist between words differing by one character. BFS guarantees the shortest path in an unweighted graph, so it's the ideal algorithm.
Use a queue to track current word and level. For each word, generate all possible one-letter variations (e.g., replace each character with 'a'-'z') and check if they are in the dictionary set for O(1) lookup.
If asked for optimization, describe bidirectional BFS: maintain two visited sets and expand the smaller frontier level by level, checking for intersection. This reduces time and space complexity.
State time complexity: O(N * L * 26) for standard BFS, where N is dictionary size and L is word length. Space: O(N). Discuss handling of start == end (return 1) and unreachable end (return 0).
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