Top 150 Interview簡単

Word Ladder

Detailed guide and Python implementation for the 'Word Ladder' problem.

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簡単

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.

Write a function ladderLength(beginWord: str, endWord: str, wordList: List[str]) -> int.

制約
  • 1 <= len(beginWord) <= 10
  • endWord.length == beginWord.length
  • 1 <= len(wordList) <= 5000
  • wordList[i].length == beginWord.length
  • beginWord, endWord, and wordList[i] consist of lowercase English letters
  • All the words in wordList are unique

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, so there is no valid transformation sequence.

Need a Hint?
Consider using Graphs-specific data structures like sets or heaps.
Edge Cases to Watch
  • Empty input structures
  • Single element inputs
  • Large numerical bounds

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