127. Word Ladder

Hard

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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.

from collections import deque
class Solution:
    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
        # Got TLE
        # Next improvement can be is to convert wordList from List to Set
        # Since check "if exists" takes linear time in List and Constant in Set
        wordList = set(wordList)
        charSet = {w for word in wordList for w in word}
        queue = deque([[beginWord, 1]])
        while queue:
            word, length = queue.popleft()
            if word == endWord:
                return length
            for index in range(len(word)):
                for char in charSet:
                    new_word = word[:index] + char + word[index+1:]
                    if new_word in wordList:
                        wordList.remove(new_word)
                        queue.append([new_word, length+1])
        return 0