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Find shortest word transformation with caching

Last updated: Mar 29, 2026

Quick Overview

This question evaluates understanding of graph algorithms for shortest-path searches, techniques for efficient neighbor generation and caching (including LRU eviction and complexity trade-offs), and concurrency controls for mutable dictionaries.

  • Medium
  • LinkedIn
  • Coding & Algorithms
  • Machine Learning Engineer

Find shortest word transformation with caching

Company: LinkedIn

Role: Machine Learning Engineer

Category: Coding & Algorithms

Difficulty: Medium

Interview Round: Technical Screen

You are given a start word and an end word of equal length, and a dictionary of valid words. In one move you may change exactly one letter to form a new word that must be in the dictionary. Find the minimal number of moves to transform the start into the end; return -1 if impossible. Implement an efficient solution. Follow-ups: ( 1) Optimize neighbor generation by introducing a reusable neighbor() function and wrap it with an LRU cache; explain eviction policy and complexity trade-offs. ( 2) The dictionary can be updated at runtime (words added/removed). Ensure correctness and thread-safety for queries while updates occur; specify how to synchronize or invalidate the neighbor cache (e.g., fine-grained locking around dictionary access or versioned snapshots) without assuming concurrent search execution.

Quick Answer: This question evaluates understanding of graph algorithms for shortest-path searches, techniques for efficient neighbor generation and caching (including LRU eviction and complexity trade-offs), and concurrency controls for mutable dictionaries.

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LinkedIn
Jul 16, 2025, 12:00 AM
Machine Learning Engineer
Technical Screen
Coding & Algorithms
8
0

You are given a start word and an end word of equal length, and a dictionary of valid words. In one move you may change exactly one letter to form a new word that must be in the dictionary. Find the minimal number of moves to transform the start into the end; return -1 if impossible. Implement an efficient solution. Follow-ups: (

  1. Optimize neighbor generation by introducing a reusable neighbor() function and wrap it with an LRU cache; explain eviction policy and complexity trade-offs. (
  2. The dictionary can be updated at runtime (words added/removed). Ensure correctness and thread-safety for queries while updates occur; specify how to synchronize or invalidate the neighbor cache (e.g., fine-grained locking around dictionary access or versioned snapshots) without assuming concurrent search execution.

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