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Word Ladder

Given two words beginWord and endWord, and a dictionary wordList, return the length of the shortest transformation sequence from beginWord to endWord, such that:

  • Only one letter can be changed at a time.

  • Each transformed word must exist in the word list.

Return 0 if there is no such transformation sequence.

Example 1:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"] 
Output: 5 
Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog", return its length 5. 

Example 2:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"] 
Output: 0 
Explanation: The endWord "cog" is not in wordList, therefore no possible transformation. 


  • 1 <= beginWord.length <= 100

  • 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 strings in wordList are unique.


class Solution {
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        Queue<String> queue = new LinkedList<String>();
        Set<String> words = new HashSet<>(wordList);
        int level = 0;
        int size = queue.size();
        for(int i=0;i<size;i++)
            String currentword = queue.poll();
            if(currentword.equals(endWord)) return level;
            List<String> neighbors = neighbor(currentword);
            for(String neigh:neighbors)
      return 0;  
    public List<String> neighbor(String s)
        char[] chars = s.toCharArray();
        List<String> result = new ArrayList<>();
        for(int i=0;i<chars.length;i++)
            char temp=chars[i];
            for(char c='a';c<='z';c++)
                String neighbor = new String(chars);
        return result;

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