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Meta·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
May 2026

Summary

Meta SWE coding round, one question on shortest word ladder. Pretty straightforward problem on the surface but the edge cases will get you if you're not careful.

Questions Asked (1)

Q1

Given two words, find the length of the shortest word ladder (transforming one word into another by changing one letter at a time, where each intermediate word must be valid).

Algorithms & Data Structures
Author's notes

Classic BFS problem and I knew that going in, but I fumbled the visited set logic at first and was computing redundant paths.

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AI HintsAI Generated

Suggested Approach

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, ensuring all intermediate words are in the given dictionary. Optimize by precomputing patterns (e.g., replacing each letter with '*') to quickly find neighbors.

Pro tip: Mention bidirectional BFS to significantly reduce search space, especially for large dictionaries, and discuss trade-offs with preprocessing time and memory. Also, clarify edge cases like when start and end words are the same or when no path exists.

1. Clarify requirements and edge cases

Confirm the dictionary is provided, words are same length, and transformations must use valid intermediate words. Discuss cases like start == end, no path, or words not in dictionary.

2. Model as a graph problem

Explain that words are nodes and edges exist between words differing by one letter. The goal is to find the shortest path from start to end.

3. Choose BFS for shortest path

Use BFS because it explores level by level, guaranteeing the shortest path in an unweighted graph. Describe how to track visited words to avoid cycles.

4. Optimize neighbor generation

Precompute a mapping from patterns (e.g., 'h*t') to words to quickly find neighbors without comparing every pair. Alternatively, generate all possible one-letter variations and check against the dictionary.

5. Analyze complexity and trade-offs

Discuss time and space complexity: O(N * L^2) for preprocessing or O(N * 26 * L) for on-the-fly generation, where N is dictionary size and L is word length. Mention bidirectional BFS as an optimization.

Key Points to Mention

  • BFS guarantees shortest path in unweighted graphs
  • Preprocessing patterns (e.g., wildcard mapping) to speed up neighbor lookup
  • Bidirectional BFS to reduce search space
  • Handling edge cases: start == end, no path, words not in dictionary
  • Time and space complexity analysis
  • Trade-offs between preprocessing and on-the-fly neighbor generation

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.