## Description

LeetCode Problem 126.

A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s_1 -> s_2 -> … -> s_k such that:

• Every adjacent pair of words differs by a single letter.
• Every s_i for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList.
• s_k == 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, s_1, s_2, …, s_k].

Example 1:

``````1
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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:

``````1
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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 <= 1000
• wordList[i].length == beginWord.length
• beginWord, endWord, and wordList[i] consist of lowercase English letters.
• beginWord != endWord
• All the words in wordList are unique.

## Sample C++ Code

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vector<vector<string>> findLadders(string beginWord, string endWord, unordered_set<string> &wordList) {
// The idea is doing BFS of paths instead of words.
vector<vector<string>> ans;
queue<vector<string>> paths;
wordList.insert(endWord);
paths.push({beginWord});
int level = 1;
int minLevel = INT_MAX;

// The "visited" records all the visited nodes on this level
// these words will never be visited again after this level
// and should be removed from wordList. This is guaranteed
// by the shortest path.
unordered_set<string> visited;

while (!paths.empty()) {
vector<string> path = paths.front();
paths.pop();
if (path.size() > level) {
//reach a new level
for (string w : visited) wordList.erase(w);
visited.clear();
if (path.size() > minLevel)
break;
else
level = path.size();
}
string last = path.back();
// Find next words in wordList by changing
// each element from 'a' to 'z'
for (int i = 0; i < last.size(); ++i) {
string news = last;
for (char c = 'a'; c <= 'z'; ++c) {
news[i] = c;
if (wordList.find(news) != wordList.end()) {
// Next word is in wordList, append this word to path.
// Path will be reused in the loop, so copy a new path.
vector<string> newpath = path;
newpath.push_back(news);
visited.insert(news);
if (news == endWord) {
minLevel = level;
ans.push_back(newpath);
}
else
paths.push(newpath);
}
}
}
}
return ans;
}
``````