Word-Ladder-BFS-127.cpp

// 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.

class Solution {
public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        unordered_set<string> wordset;
        bool exists = false;
        
        for(auto word:wordList){
            if(word == endWord)
                exists = true;
            wordset.insert(word);
        }
        if(!exists) return 0;
        
        queue<string> q;
        q.push(beginWord);
        int level = 0;
        
        while(!q.empty()){
            level++;
            int len = q.size();
            while(len--){
                string current = q.front();
                q.pop();
                for(int i = 0; i < current.size(); i++){
                    string temp = current;
                    for(char c = 'a'; c <= 'z'; c++){
                        temp[i] = c;
                        if(temp == current) continue;
                        if(temp == endWord)
                            return level+1;
                        if(wordset.find(temp) != wordset.end()){
                            q.push(temp);
                            wordset.erase(temp);
                        }
                    }
                }
            }
        }
        return 0;
    }
};