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Find shortest word transformation with caching

Quick Overview

Find shortest word transformation with caching evaluates algorithm design, data structures, correctness, complexity, edge cases, and implementation details in a realistic interview setting. A strong answer states assumptions, handles edge cases, explains trade-offs, and shows how to validate the result clearly.

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: Find shortest word transformation with caching evaluates algorithm design, data structures, correctness, complexity, edge cases, and implementation details in a realistic interview setting. A strong answer states assumptions, handles edge cases, explains trade-offs, and shows how to validate the result clearly.

You are given a start word `begin_word`, an end word `end_word` of equal length, and a dictionary `word_list` of valid words. In one move you may change exactly one letter of the current word to form a new word, and that new word must be present in the dictionary. Return the minimal number of moves required to transform `begin_word` into `end_word`. If the transformation is impossible, return -1. Notes / assumptions: - A "move" is a single letter change, so the answer is the number of edges (transformations), not the number of words visited. - If `begin_word == end_word`, the answer is 0 (no moves needed). - `end_word` must itself be in the dictionary to be reachable; otherwise return -1. - `begin_word` is the source and need not be in the dictionary. Follow-ups to discuss (not graded by the tests): (1) Factor neighbor generation into a reusable `neighbor()` function wrapped with an LRU cache; discuss eviction policy and the time/space trade-off. (2) Handle a dictionary that can be updated at runtime (words added/removed): describe how to keep queries correct and how to synchronize or invalidate the neighbor cache (e.g. fine-grained locking around dictionary access, or versioned snapshots) without assuming concurrent search execution.

Constraints

1 <= len(begin_word) == len(end_word) == L
All words in word_list have length L.
0 <= len(word_list)
Words consist of lowercase English letters.
end_word may or may not be present in word_list (if absent, return -1).

Examples

Input: ("hit", "cog", ["hot", "dot", "dog", "lot", "log", "cog"])

Expected Output: 4

Explanation: hit -> hot -> dot -> dog -> cog is the shortest path: 4 moves.

Input: ("hit", "cog", ["hot", "dot", "dog", "lot", "log"])

Expected Output: -1

Explanation: end_word 'cog' is not in the dictionary, so the transformation is impossible.

Input: ("a", "c", ["a", "b", "c"])

Expected Output: 1

Explanation: Single-letter words: a -> c directly in one move since 'c' is in the dictionary.

Input: ("hot", "hot", ["hot", "dot", "dog"])

Expected Output: 0

Explanation: begin_word already equals end_word, so zero moves are needed.

Input: ("hit", "cog", [])

Expected Output: -1

Explanation: Empty dictionary: end_word is unreachable, return -1.

Input: ("red", "tax", ["ted", "tex", "red", "tax", "tad", "den", "rex", "pee"])

Expected Output: 3

Explanation: red -> ted -> tad -> tax is the shortest path: 3 moves.

Hints

Model words as graph nodes; connect two words with an edge if they differ by exactly one letter. The shortest transformation is the shortest path, so BFS gives the minimal number of moves.
Generate neighbors of a word by trying all 25 other letters at each of the L positions and keeping only candidates that exist in the dictionary set.
Track visited words to avoid revisiting; the first time you dequeue end_word, the accumulated move count is optimal.
Edge cases: begin_word == end_word returns 0; end_word not in the dictionary returns -1; an empty dictionary returns -1.

Quick Overview