Model the problem as a graph where each word is a node and edges connect words that differ by one character. Use BFS to find the shortest path from the start word to the end word, ensuring all intermediate words are in the given word list. If the end word is not reachable, return -1.
Pro tip: Mention that bidirectional BFS can significantly reduce the search space, especially for large word lists, and discuss the trade-offs between preprocessing (e.g., building adjacency lists) and on-the-fly neighbor generation.
Confirm assumptions: Are all words the same length? Is the transformation case-sensitive? Can the start or end word be absent from the word list? Discuss edge cases like start == end.
Represent each word as a node. Two nodes are connected if they differ by exactly one character. The word list defines the set of valid nodes.
Since each transformation has unit cost, BFS guarantees the minimum number of steps. Initialize a queue with the start word and track visited words to avoid cycles.
Instead of comparing all pairs, generate neighbors by changing each character to 'a'-'z' and checking if the result is in the word set. Alternatively, use a precomputed pattern map (e.g., '*ot' -> [hot, lot]).
Time: O(N * L * 26) where N is word list size and L is word length. Space: O(N). Discuss bidirectional BFS to reduce time to O(b^(d/2)) and when preprocessing is beneficial.
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