Word Ladder

Problem Statement

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.

Leetcode Link

Examples:

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.

Code

Python Code

class Solution:
    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
        s=set(wordList)
        l=['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t',
        'u','v','w','x','y','z']
        queue=deque([])
        queue.append([beginWord,0])
        while queue:
            a,b=queue.popleft()
            if a==endWord:
                return b+1
            for j in range(len(a)):
                for i in l:
                    if (a[:j]+i+a[j+1:]) in s and (a[:j]+i+a[j+1:])!=beginWord:
                        s.remove(a[:j]+i+a[j+1:])
                        queue.append([a[:j]+i+a[j+1:],b+1])
        return 0