Model the problem as finding the shortest path in an unweighted graph where nodes are dictionary words and edges connect words that differ by one character. Use BFS from the start word to the end word, ensuring each intermediate word is in the dictionary. If the end word is not in the dictionary, return 0 or -1 as appropriate.
Pro tip: Preprocess the dictionary into a set for O(1) lookups, and generate neighbors by trying all 26 lowercase letters at each position. This avoids O(N^2) pairwise comparisons and is the expected efficient solution.
Confirm that each operation changes exactly one character, and that all intermediate words must be in the dictionary. Ask about cases where start or end word is not in the dictionary, or when no transformation is possible.
Treat each dictionary word as a node, with edges between words that differ by one character. The goal is to find the shortest path from start to end, which corresponds to the minimum number of operations.
Since edges are unweighted, BFS guarantees the shortest path. Initialize a queue with the start word and a visited set 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 dictionary set. This reduces time complexity to O(L * 26 * N) where L is word length and N is dictionary size.
State time and space complexity, and mention alternative approaches like bidirectional BFS for further optimization. Discuss handling of large dictionaries and potential memory constraints.
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