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Bytedance·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
May 2026

Summary

Bytedance SWE interview that leaned hard into graph/BFS problems. The Word Ladder family of questions showed up and they wanted more than just the length, they wanted actual path reconstruction too.

Questions Asked (1)

Q1

Given a start word, an end word, and a list of equal-length words, find the length of the shortest transformation sequence where each step changes exactly one letter and every intermediate word must appear in the word list. Then extend the solution to return the actual shortest path (or all shortest paths), not just the length.

Algorithms & Data Structures
Author's notes

I got the BFS part for the length pretty quickly, that felt fine.

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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 to find the shortest transformation length, then extend BFS to track predecessors for path reconstruction or to collect all shortest paths.

Pro tip: When extending to all shortest paths, avoid exponential blowup by storing only the shortest distance to each word and building paths backward from the end word using the distance map.

1. Clarify and Model

Confirm constraints (word length, list size, case sensitivity) and model the problem as an unweighted graph where nodes are words and edges connect words differing by exactly one letter.

2. BFS for Shortest Length

Run BFS from the start word, exploring neighbors level by level. Stop when the end word is reached, returning the current depth as the shortest length.

3. Optimize Neighbor Generation

Use precomputed patterns (e.g., replacing each character with '*') to group words and quickly find neighbors, reducing time from O(N^2 * L) to O(N * L).

4. Extend to Path Reconstruction

During BFS, store each word's predecessor(s) or distance. After reaching the end, backtrack from the end word to the start using the stored information to reconstruct one or all shortest paths.

5. Handle All Shortest Paths

To return all shortest paths, maintain a list of predecessors for each word at the same BFS level. Then perform DFS from the end word to collect all paths, ensuring no cycles.

Key Points to Mention

  • BFS guarantees shortest path in unweighted graphs.
  • Time complexity: O(N * L) with pattern optimization, where N is number of words and L is word length.
  • Space complexity: O(N * L) for storing patterns and predecessors.
  • Bidirectional BFS can reduce search space and improve performance.
  • For all shortest paths, avoid revisiting nodes at the same or greater depth to prevent cycles and redundant work.
  • Edge cases: start equals end, no path exists, duplicate words in list.

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