Word ladder (shortest path)

qwen3-30b-a3b: timed out

Create ladder.py with exactly one function:

ladder_length(begin: str, end: str, word_list: list[str]) -> int
    A transformation sequence goes from begin to end, changing exactly one letter per step; every intermediate word (and end itself) must appear in word_list. Return the number of words in the SHORTEST such sequence counting both endpoints, or 0 if none exists. begin does not need to be in word_list. All words are lowercase and the same length.

Example: begin="hit", end="cog", word_list=["hot","dot","dog","lot","log","cog"] -> 5 (hit -> hot -> dot -> dog -> cog).

Use only the Python standard library. A breadth-first search is the intended approach; brute-force recursion will be too slow for the larger test.

import time from ladder import ladder_length  def test_classic_example():    assert ladder_length("hit", "cog", ["hot", "dot", "dog", "lot", "log", "cog"]) == 5  def test_unreachable_end():    assert ladder_length("hit", "cog", ["hot", "dot", "dog", "lot", "log"]) == 0  def test_single_step():    assert ladder_length("a", "c", ["a", "b", "c"]) == 2  def test_shortest_is_found():    words = ["hot", "hit", "hat", "cat", "cot", "cog", "dog"]    assert ladder_length("hit", "cot", words) == 3  def test_larger_input_is_fast():    words = [a + b + c for a in "abcdefgh" for b in "abcdefgh" for c in "abcdefgh"]    start = time.monotonic()    assert ladder_length("aaa", "hhh", words) == 4    assert time.monotonic() - start < 2.0

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