Find shortest word transformation with caching

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.

What a Strong Answer Covers

• A clear algorithm with the right data structures and enough pseudocode or code-level detail to implement it.
• A correctness argument that explains why the algorithm covers all required cases.
• Time and space complexity, plus at least one alternative approach when relevant.
• Focused tests for normal cases, edge cases, and failure modes.

Follow-up Questions

• How would the approach change if the input were streaming or too large for memory?
• What invariants would you assert in production code?
• Which tests would catch off-by-one, duplicate, or tie-breaking bugs?

Shortest Word Transformation (Word Ladder)

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

Example 1

Input: ("hit", "cog", ["hot", "dot", "dog", "lot", "log", "cog"])
Output: 4

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

Example 2

Input: ("hit", "cog", ["hot", "dot", "dog", "lot", "log"])
Output: -1

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

Hints

1. 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.
2. 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.
3. Track visited words to avoid revisiting; the first time you dequeue end_word, the accumulated move count is optimal.
4. Edge cases: begin_word == end_word returns 0; end_word not in the dictionary returns -1; an empty dictionary returns -1.

Solution

Solution Alignment

The prompt asks for an implementation-level answer. The safest way to present it is to define the state, maintain clear invariants, then walk through complexity and tests.

Problem Restatement

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., fi...

Recommended Approach

Use a hash map from key to doubly linked-list node plus a doubly linked list ordered by recency. get moves the node to the front. put updates and moves an existing node, or inserts a new node at the front and evicts the tail when capacity is exceeded.

Correctness

The implementation should maintain an invariant after each loop or operation that directly matches the problem statement. At termination, that invariant implies the returned value has considered every valid candidate exactly once, or has preserved the required data-structure state after every API call.

Complexity

get and put are O(1) average time. Space is O(capacity).

Edge Cases and Tests

Capacity 0 or 1, updating an existing key, eviction order after get, repeated puts, and missing-key gets.