127. Word Ladder

Description

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.

 

Solutions

Solution: Breadth-First Search

• Time complexity: O(n*26^word.length)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {string} beginWord
 * @param {string} endWord
 * @param {string[]} wordList
 * @return {number}
 */
const ladderLength = function (beginWord, endWord, wordList) {
  if (!wordList.includes(endWord)) return 0;
  const BASE_CHAR_CODE = 'a'.charCodeAt(0);
  const wordSet = new Set(wordList);
  let queue = [beginWord];
  let result = 1;

  while (queue.length) {
    const nextQueue = [];

    for (const word of queue) {
      for (let index = 0; index < word.length; index++) {
        for (let code = BASE_CHAR_CODE; code < BASE_CHAR_CODE + 26; code++) {
          const s = String.fromCharCode(code);
          const targetWord = `${word.slice(0, index)}${s}${word.slice(index + 1)}`;

          if (!wordSet.has(targetWord)) continue;
          if (targetWord === endWord) return result + 1;
          wordSet.delete(targetWord);
          nextQueue.push(targetWord);
        }
      }
    }
    queue = nextQueue;
    result += 1;
  }
  return 0;
};