126. Word Ladder II

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 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.
• The sum of all shortest transformation sequences does not exceed 105.

 

Solutions

Solution: Breadth-First Search

• Time complexity: O(n*word.length²⁶)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {string} beginWord
 * @param {string} endWord
 * @param {string[]} wordList
 * @return {string[][]}
 */
const findLadders = function (beginWord, endWord, wordList) {
  const BASE_CHAR_CODE = 'a'.charCodeAt(0);
  const wordSet = new Set(wordList);

  if (!wordSet.has(endWord)) return [];
  const sequences = [];
  let queue = [beginWord];
  let reached = false;

  wordSet.delete(beginWord);

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

    sequences.push(queue);
    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 letter = String.fromCharCode(code);
          const nextWord = `${word.slice(0, index)}${letter}${word.slice(index + 1)}`;

          if (!wordSet.has(nextWord)) continue;
          if (nextWord === endWord) {
            reached = true;
            break;
          }
          nextQueue.push(nextWord);
          wordSet.delete(nextWord);
        }
        if (reached) break;
      }
    }
    queue = nextQueue;
  }
  if (!reached) return [];

  const result = [[endWord]];
  const isValid = (a, b) => {
    let diff = 0;

    for (const [index, element] of a.entries()) {
      if (element !== b[index]) diff += 1;
      if (diff > 1) return false;
    }
    return diff === 1;
  };

  for (let step = sequences.length - 1; step >= 0; step--) {
    const size = result.length;

    for (let index = 0; index < size; index++) {
      const path = result.shift();
      const nextWord = path[0];

      for (const word of sequences[step]) {
        if (!isValid(word, nextWord)) continue;
        result.push([word, ...path]);
      }
    }
  }
  return result;
};