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[8/14] 126. Word Ladder II

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작성일 2022.08.20 21:46

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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 all the shortest transformation sequences from beginWord to endWord, or an empty list if no such sequence exists. Each sequence should be returned as a list of the words [beginWord, s1, s2, ..., sk].

Example 1:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
Output: [["hit","hot","dot","dog","cog"],["hit","hot","lot","log","cog"]]
Explanation: There are 2 shortest transformation sequences:
"hit" -> "hot" -> "dot" -> "dog" -> "cog"
"hit" -> "hot" -> "lot" -> "log" -> "cog"

Example 2:

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

Constraints:

1 <= beginWord.length <= 5
endWord.length == beginWord.length
1 <= wordList.length <= 500
wordList[i].length == beginWord.length
beginWord, endWord, and wordList[i] consist of lowercase English letters.
beginWord != endWord
All the words in wordList are unique.

Accepted

326,866

Submissions

1,183,724

#Amazon, #Microsoft, #Bloomberg, #Expedia, #Dunzo

링크

https://leetcode.com/problems/word-ladder-ii/ 4318 회 연결

익명
작성일 2022.08.20 21:47

class Solution:
    def findLadders(self, beginWord: str, endWord: str, wordList: List[str]) -> List[List[str]]:
        wordList = set(wordList)
        if endWord not in wordList:
            return []
        # BFS visit
        curr_level = {beginWord}
        parents = collections.defaultdict(list)
        while curr_level:
            wordList -= curr_level
            next_level = set()
            for word in curr_level:
                for i in range(len(word)):
                    for c in 'abcdefghijklmnopqrstuvwxyz':
                        new_word = word[:i] + c + word[i+1:]
                        if new_word in wordList:
                            next_level.add(new_word)
                            parents[new_word].append(word)
            if endWord in next_level:
                break
            curr_level = next_level
        # DFS reconstruction
        res = []
        def dfs(word, path):
            if word == beginWord:
                path.append(word)
                res.append(path[::-1])
            else:
                for p_word in parents[word]:
                    dfs(p_word, path + [word])
        dfs(endWord, [])
        return res

[code=Python 3]class Solution:
    def findLadders(self, beginWord: str, endWord: str, wordList: List[str]) -> List[List[str]]:
        wordList = set(wordList)
        if endWord not in wordList:
            return []
        # BFS visit
        curr_level = {beginWord}
        parents = collections.defaultdict(list)
        while curr_level:
            wordList -= curr_level
            next_level = set()
            for word in curr_level:
                for i in range(len(word)):
                    for c in 'abcdefghijklmnopqrstuvwxyz':
                        new_word = word[:i] + c + word[i+1:]
                        if new_word in wordList:
                            next_level.add(new_word)
                            parents[new_word].append(word)
            if endWord in next_level:
                break
            curr_level = next_level
        # DFS reconstruction
        res = []
        def dfs(word, path):
            if word == beginWord:
                path.append(word)
                res.append(path[::-1])
            else:
                for p_word in parents[word]:
                    dfs(p_word, path + [word])
        dfs(endWord, [])
        return res[/code]