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Snapchat·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Jul 2026

Summary

Snapchat software engineer technical phone screen. One graph/BFS-style problem, pretty standard stuff but the exact boolean-only constraint tripped me up a little because I kept wanting to return the path length out of habit.

Questions Asked (1)

Q1

Given a start word, an end word, and a dictionary of valid words, determine whether it's possible to transform the start word into the end word by changing exactly one letter at a time, where every intermediate word must exist in the dictionary. Return true or false.

Algorithms & Data Structures
Author's notes

My first instinct was BFS and that was correct, but I wasted a few minutes mentally solving for shortest path length before re-reading and realizing they just wanted a boolean.

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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 from the start word to find the shortest path to the end word, checking only words in the dictionary. Return true if the end word is reached, else false.

Pro tip: Optimize by using bidirectional BFS to reduce search space, and preprocess the dictionary into patterns (e.g., replacing each letter with '*') to quickly find neighbors. This demonstrates strong problem-solving skills and efficiency awareness.

1. Clarify and Validate Input

Confirm edge cases: start and end words may be the same, dictionary may not contain start or end, and all words are of equal length. Discuss assumptions with the interviewer.

2. Model as Graph

Treat each word as a node. Two words are connected if they differ by exactly one letter. The dictionary defines the set of valid nodes.

3. Choose BFS for Shortest Path

Use BFS to explore level by level, ensuring the shortest transformation sequence. This is optimal for unweighted graphs.

4. Optimize Neighbor Generation

Preprocess the dictionary into a map from pattern (e.g., '*ot') to list of words. This allows O(1) neighbor lookup per pattern.

5. Implement and Analyze Complexity

Code the BFS with visited set. Analyze time complexity: O(N * L^2) where N is dictionary size and L is word length, or O(N * L) with pattern map. Space complexity O(N).

Key Points to Mention

  • Graph modeling: words as nodes, edges for one-letter differences.
  • BFS guarantees shortest path in unweighted graphs.
  • Bidirectional BFS can significantly reduce search space.
  • Preprocessing dictionary with wildcard patterns for efficient neighbor lookup.
  • Handling edge cases: start == end, missing words in dictionary.
  • Time and space complexity analysis.

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