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:

1. Only one letter can be changed at a time.
2. 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.

class Solution {
public:
int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
unordered_set<string> wordSet(wordList.begin(), wordList.end());
if (!wordSet.count(endWord)) return 0;
unordered_map<string, int> pathCnt{{{beginWord, 1}}};
queue<string> q{{beginWord}};
while (!q.empty()) {
string word = q.front(); q.pop();
for (int i = 0; i < word.size(); ++i) {
string newWord = word;
for (char ch = 'a'; ch <= 'z'; ++ch) {
newWord[i] = ch;
if (wordSet.count(newWord) && newWord == endWord) return pathCnt[word] + 1;
if (wordSet.count(newWord) && !pathCnt.count(newWord)) {
q.push(newWord);
pathCnt[newWord] = pathCnt[word] + 1;
}
}
}
}
return 0;
}
};

class Solution {
public:
int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
unordered_set<string> wordSet(wordList.begin(), wordList.end());
if (!wordSet.count(endWord)) return 0;
queue<string> q{{beginWord}};
int res = 0;
while (!q.empty()) {
for (int k = q.size(); k > 0; --k) {
string word = q.front(); q.pop();
if (word == endWord) return res + 1;
for (int i = 0; i < word.size(); ++i) {
string newWord = word;
for (char ch = 'a'; ch <= 'z'; ++ch) {
newWord[i] = ch;
if (wordSet.count(newWord) && newWord != word) {
q.push(newWord);
wordSet.erase(newWord);
}
}
}
}
++res;
}
return 0;
}
};

Minimum Genetic Mutation