Word Ladder (BFS)

To solve the Word Ladder problem, I first model the dictionary as an unweighted graph where each word is a node. Since we need the shortest transformation sequence, Breadth-First Search is the optimal choice because it explores layer by layer, guaranteeing the first time we reach the endWord, it is via the minimum number of steps. My approach involves initializing a queue with the beginWord and a visited set containing it. In each iteration, I dequeue the current word and generate all possible neighbors by iterating through each character position and replacing it with letters 'a' through 'z'. If a generated word exists in the dictionary and hasn't been visited, I add it to the queue and mark it as visited. This avoids the inefficiency of checking the entire dictionary for every word. For example, transforming 'hit' to 'cog' with a dictionary including 'hot', 'dot', 'dog', 'lot', 'log', and 'cog' would proceed: hit -> hot -> dot/dog -> lot/log -> cog. The algorithm returns 5 steps. I would ensure to handle edge cases like missing endWords or unreachable paths by returning zero. This solution runs in O(M^2 * N) time, which aligns well with Meta's expectation for efficient, scalable graph algorithms in production systems.