LC 127 - Word Ladder

Question

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.

Links

Question here and solution here

Solution

concept

The key concept is to realise this is a graph problem. The most difficult part of the question is to construct the adjacency list. The normal way of doing it will cause TLE.

code

class Solution:
    """
    graph + BFS
    each node is a word, each edge is a transformation (single edit distance)
    find the shortest path from beginWord to endWord
    """
    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
        if endWord not in wordList:
            return 0
        # create adjacency list: {pattern: [word]}
        # e.g. {*it : [hit, hot]}
        # use a wild card pattern such that this process is n*m^2
        # if use double loop is is n^2*m which will TLE
        neighbor_map = defaultdict(list)
        wordList.append(beginWord) # wordList does not contain begin word
        for word in wordList:
            for j in range(len(word)):
                pattern = word[:j] + "*" + word[j+1:]
                neighbor_map[pattern].append(word) 
        # BFS
        visited = set([beginWord])
        q = deque([beginWord])
        result = 1

        while q:
            for i in range(len(q)):
                word = q.popleft()
                if word == endWord:
                    return result
                for j in range(len(word)):
                    pattern = word[:j] + "*" + word[j+1:]
                    for neighbor in neighbor_map[pattern]:
                        if neighbor not in visited:
                            visited.add(neighbor)
                            q.append(neighbor)
            result += 1

        return 0

Complexity

time: $O(n^2 \times m)$
space: $O(n^2)$