Word Ladder (BFS)
Given two words, `beginWord` and `endWord`, and a dictionary, find the length of the shortest transformation sequence from `beginWord` to `endWord` (one letter change per step). Model the dictionary as a graph and use BFS.
Why Interviewers Ask This
Meta asks this to evaluate a candidate's ability to model real-world problems as unweighted graphs and implement Breadth-First Search efficiently. It tests spatial reasoning, understanding of queue-based traversal, and the capacity to optimize for time complexity in constrained environments where shortest path logic is critical.
How to Answer This Question
1. Clarify constraints immediately: Ask if the dictionary is case-sensitive or if beginWord equals endWord.
2. Model the problem: Explicitly state that words are nodes and single-letter differences represent edges.
3. Choose BFS: Explain why BFS guarantees the shortest path in an unweighted graph compared to DFS.
4. Optimize neighbor generation: Propose generating valid next words by changing one character at a time rather than scanning the entire dictionary every step.
5. Implement with a queue: Describe using a deque for the current level and a set for visited nodes to prevent cycles and redundant processing.
6. Discuss complexity: Analyze O(M^2 * N) time where M is word length and N is dictionary size, highlighting Meta's focus on scalable solutions.
Key Points to Cover
- Explicitly identifying the problem as an unweighted graph search
- Justifying BFS over DFS for finding the shortest path
- Optimizing neighbor generation to avoid O(N) dictionary scans per node
- Using a Set for O(1) lookups to track visited states and prevent cycles
- Clearly articulating time complexity relative to word length and dictionary size
Sample Answer
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 e…
Common Mistakes to Avoid
- Using Depth-First Search (DFS) which fails to guarantee the shortest path
- Scanning the entire dictionary for every word instead of generating neighbors
- Failing to track visited nodes, leading to infinite loops in cyclic paths
- Ignoring edge cases where the endWord is not present in the dictionary
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