---
title: '127. Word Ladder'
description: 'A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> ... -> sk such that:'
sidebar:
  label: 'Word Ladder'
  badge: 'Hard'
---

Hash Table

### Example 1:
- Input: `beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]`
- Output: `5`
- Explanation: One shortest transformation sequence is "hit" -> "hot" -> "dot" -> "dog" -> cog", which is `5` words long.

### Example 2:
- Input: `beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]`
- Output: `0`
- Explanation: The `endWord` "cog" is not in `wordList`, therefore there is no valid transformation sequence.

### Constraints:

- `1 <= beginWord.length <= 10`
- `endWord.length == beginWord.length`
- `1 <= wordList.length <= 5000`
- `wordList[i].length == beginWord.length`
- `beginWord`, `endWord`, and `wordList[i]` consist of lowercase English letters.
- `beginWord != endWord`
- All the words in `wordList` are unique.

## Solution

```py
import collections


class Solution:
    def ladderLength(self, beginWord: str, endWord: str, wordList: list[str]) -> int:
        if endWord not in wordList:
            return 0

        adj = collections.defaultdict(list)
        wordList.append(beginWord)

        for word in wordList:
            for j in range(len(word)):
                pattern = word[:j] + "*" + word[j + 1 :]
                adj[pattern].append(word)

        visit = set([beginWord])
        q = collections.deque([beginWord])
        res = 1

        while q:
            for i in range(len(q)):
                word = q.popleft()
                if word == endWord:
                    return res
                for j in range(len(word)):
                    pattern = word[:j] + "*" + word[j + 1 :]
                    for adjWord in adj[pattern]:
                        if adjWord not in visit:
                            visit.add(adjWord)
                            q.append(adjWord)
            res += 1
        return 0
```