Hard
Word Ladder
Hard
0 submissions
50 coins
+200 XP
Breadth-First Search
Hash Table
String
Problem Description
## Problem
A **transformation sequence** from word `beginWord` to word `endWord` using a dictionary `wordList` is a sequence of words `beginWord -> s1 -> s2 -> ... -> sk` such that:
- Every adjacent pair of words differs by a single letter.
- Every `si` for `1 <= i <= k` is in `wordList`. Note that `beginWord` does not need to be in `wordList`.
- `sk == endWord`
Given two words `beginWord` and `endWord`, and a dictionary `wordList`, return the **number of words in the shortest transformation sequence** from `beginWord` to `endWord`, or `0` if no such sequence exists.
## Examples
**Example 1:**
```
Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
Output: 5
Explanation: hit -> hot -> dot -> dog -> cog (5 words in the sequence)
```
**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.
```
**Example 3:**
```
Input: beginWord = "a", endWord = "c", wordList = ["a","b","c"]
Output: 2
Explanation: a -> c (single letter change, 2 words in sequence)
```
## Approach Hints
Use BFS. Start from `beginWord` and explore all one-letter transformations. Use a set for `wordList` to check membership in O(1). Track visited words to avoid cycles. The BFS level when `endWord` is found is the answer (add 1 for `beginWord`).
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 words in `wordList` are unique.
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