127. Word Ladder

Problem

Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that:

• Only one letter can be changed at a time.
• Each transformed word must exist in the word list. Note that beginWord is not a transformed word.

Note:

• Return 0 if there is no such transformation sequence.
• All words have the same length.
• All words contain only lowercase alphabetic characters.
• You may assume no duplicates in the word list.
• You may assume beginWord and endWord are non-empty and are not the same.

Example 1:

Input:
beginWord = "hit",
endWord = "cog",
wordList = ["hot","dot","dog","lot","log","cog"]

Output: 5

Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Example 2:

Input:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

Output: 0

Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.

Solution (Java)

class Solution {
    public int ladderLength(String beginWord, String endWord, List<String> wordDict) {
        Set<String> beginSet = new HashSet<>();
        Set<String> endSet = new HashSet<>();
        Set<String> wordSet = new HashSet<>(wordDict);
        Set<String> visited = new HashSet<>();
        if (!wordDict.contains(endWord)) {
            return 0;
        }
        int len = 1;
        int strLen = beginWord.length();
        beginSet.add(beginWord);
        endSet.add(endWord);
        while (!beginSet.isEmpty() && !endSet.isEmpty()) {
            if (beginSet.size() > endSet.size()) {
                Set<String> temp = beginSet;
                beginSet = endSet;
                endSet = temp;
            }
            Set<String> tempSet = new HashSet<>();
            for (String s : beginSet) {
                char[] chars = s.toCharArray();
                for (int i = 0; i < strLen; i++) {
                    char old = chars[i];
                    for (char j = 'a'; j <= 'z'; j++) {
                        chars[i] = j;
                        String temp = new String(chars);
                        if (endSet.contains(temp)) {
                            return len + 1;
                        }
                        if (!visited.contains(temp) && wordSet.contains(temp)) {
                            tempSet.add(temp);
                            visited.add(temp);
                        }
                    }
                    chars[i] = old;
                }
            }
            beginSet = tempSet;
            len++;
        }
        return 0;
    }
}

Solution (Javascript)

/**
 * @param {string} beginWord
 * @param {string} endWord
 * @param {string[]} wordList
 * @return {number}
 */
var ladderLength = function(beginWord, endWord, wordList) {
  var wordSet = new Set(wordList);
  var queue = [];
  var step = 0;
  var word = '';
  var len = 0;
  var i = 0;

  pushNextWord(beginWord, queue, wordSet);
  step = 2;

  while (len = queue.length) {
    for (i = 0; i < len; i++) {
      word = queue.shift();
      if (word === endWord) return step;
      pushNextWord(word, queue, wordSet);
    }
    step++;
  }

  return 0;
};

var pushNextWord = function (word, queue, wordSet) {
  var start = 'a'.charCodeAt(0);
  var len = word.length;
  var str = '';

  wordSet.delete(word);

  for (var i = 0; i < len; i++) {
    for (var j = 0; j < 26; j++) {
      str = word.substr(0, i) + String.fromCharCode(j + start) + word.substr(i + 1);

      if (wordSet.has(str)) {
        queue.push(str);
        wordSet.delete(str);
      }
    }
  }
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).