Word Ladder

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.

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:

class Solution:
    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
        n = len(beginWord)
        wordList = set(wordList + [beginWord])
        graph = {}
        for word in wordList:
            for i in range(n):
                for c in "abcdefghijklmnopqrstuvwxyz":
                    currstr = word[0:i] + c + word[i+1:]
                    if currstr != word and currstr in wordList:
                        if word in graph:
                            graph[word].append(currstr)
                        else:
                            graph[word] = [currstr]
        dist = {}
        dist[beginWord] = 1
        queue = [(beginWord, 1)]
        visited = set()
        i = 0
        while i < len(queue):
            curr, k = queue[i]
            for j in graph.get(curr, []):
                if j not in visited:
                    dist[j] = dist.get(curr, float('inf')) + 1
                    if j == endWord:
                        return k + 1
                    visited.add(j)
                    queue.append((j, k + 1))
            i += 1
        return 0