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November 12, 2022 06:05 pm GMT

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.

class Solution:    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:        s = set(wordList)        l = list("abcdefghijklmnopqrstuvwxyz")        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

Original Link: https://dev.to/salahelhossiny/word-ladder-4o0d

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