Meta·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Apr 2026

Summary

Meta SWE coding round, one graph/string problem the whole session. Pretty standard setup but the problem had enough wrinkles to keep it interesting.

Questions Asked (1)

Given a start word, an end word, and a dictionary of valid words, find any path from the start word to the end word by changing one letter at a time, where each intermediate word must exist in the dictionary. The path doesn't need to be the shortest.

Algorithms & Data Structures

Author's notes

The 'doesn't need to be shortest' part threw me off a little because my brain immediately went to BFS since that's the classic word ladder setup.

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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 that differ by one letter. Use BFS to explore from the start word, tracking the path, until the end word is found. Since any path is acceptable, BFS will find a path (not necessarily shortest) efficiently.

Pro tip: Preprocess the dictionary into a set for O(1) lookups, and generate neighbors by trying all 26 letter substitutions per position. This avoids comparing against every word in the dictionary, which is crucial for large inputs.

1. Clarify and Validate

Confirm assumptions: dictionary includes start and end words? Case sensitivity? Can we reuse words? Ensure the problem is well-defined before proceeding.

2. Model as Graph

Treat each word as a node. Two words are connected if they differ by exactly one letter. The goal is to find any path from start to end.

3. Choose BFS and Optimize Neighbor Generation

Use BFS for pathfinding. For each word, generate neighbors by changing each character to 'a'-'z' and checking if the result is in the dictionary set.

4. Track Path and Avoid Cycles

Maintain a visited set to avoid revisiting words. Store parent pointers or the path itself to reconstruct the sequence once the end word is reached.

5. Analyze Complexity and Edge Cases

Discuss time complexity: O(N * L * 26) where N is dictionary size and L is word length. Handle cases like no path exists, start equals end, or words of different lengths.

Key Points to Mention

Graph representation: words as nodes, edges for one-letter differences.
BFS guarantees a path if one exists, and is optimal for unweighted graphs.
Efficient neighbor generation using a set and trying all 26 letters per position.
Use a visited set to prevent cycles and redundant work.
Path reconstruction using parent pointers or storing paths in the queue.
Time and space complexity analysis, and handling edge cases like no path.

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