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94 lines
2.5 KiB
Python
94 lines
2.5 KiB
Python
"""
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127. Word Ladder
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Difficulty: Hard
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https://leetcode.com/problems/word-ladder/
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──────────────────────────────────────────────────
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A transformation sequence from word beginWord to word endWord using a
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dictionary wordList is a sequence of words beginWord -> s1 -> s2 ->
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... -> sk such that:
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• Every adjacent pair of words differs by a single letter.
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• Every si for 1 <= i <= k is in wordList. Note that beginWord does
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not need to be in wordList.
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• sk == endWord
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Given two words, beginWord and endWord, and a dictionary wordList,
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return the number of words in the shortest transformation sequence
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from beginWord to endWord, or 0 if no such sequence exists.
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Example 1:
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Input: beginWord = "hit", endWord = "cog", wordList =
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["hot","dot","dog","lot","log","cog"]
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Output: 5
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Explanation: One shortest transformation sequence is "hit" -> "hot"
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-> "dot" -> "dog" -> cog", which is 5 words long.
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Example 2:
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Input: beginWord = "hit", endWord = "cog", wordList =
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["hot","dot","dog","lot","log"]
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Output: 0
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Explanation: The endWord "cog" is not in wordList, therefore there is
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no valid transformation sequence.
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Constraints:
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• 1 <= beginWord.length <= 10
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• endWord.length == beginWord.length
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• 1 <= wordList.length <= 5000
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• wordList[i].length == beginWord.length
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• beginWord, endWord, and wordList[i] consist of lowercase English
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letters.
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• beginWord != endWord
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• All the words in wordList are unique.
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"""
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import collections
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class Solution:
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def ladderLength(self, beginWord: str, endWord: str, wordList: list[str]) -> int:
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if endWord not in wordList:
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return 0
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adj = collections.defaultdict(list)
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wordList.append(beginWord)
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for word in wordList:
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for j in range(len(word)):
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pattern = word[:j] + "*" + word[j + 1 :]
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adj[pattern].append(word)
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visit = set([beginWord])
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q = collections.deque([beginWord])
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res = 1
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while q:
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for i in range(len(q)):
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word = q.popleft()
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if word == endWord:
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return res
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for j in range(len(word)):
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pattern = word[:j] + "*" + word[j + 1 :]
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for adjWord in adj[pattern]:
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if adjWord not in visit:
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visit.add(adjWord)
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q.append(adjWord)
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res += 1
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return 0
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