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.

Code

Time Complexity: $O(NM)$, Space Complexity: $O(N)$, where $M$ is the length of a single word.

class Solution {
public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        unordered_set<string> s(wordList.begin(), wordList.end());
        queue<string> q;
        q.push(beginWord);
        int ladder = 1;
        while(!q.empty()) {
            int n = q.size();
            for(int k = 0; k < n; k++) {
                string word = q.front();
                q.pop();
                if(word == endWord) return ladder;
                // prevent infinite loop
                s.erase(word);
                for(int i = 0; i < word.length(); i++) {
                    char c = word[i];
                    for(int j = 0; j < 26; j++) {
                        word[i] = 'a' + j;
                        if(s.find(word) != s.end()) {
                            q.push(word);
                        }
                    }
                    word[i] = c;
                }
            }
 
            ladder++;
        }
        return 0;
    }
};

Source

• Word Ladder - LeetCode