LeetCode #127

# Description:

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:

Only one letter can be changed at a time. 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:

``````Given:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log","cog"]
As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.
``````

# Code:

``````class Solution {
public:
int ladderLength(string beginWord, string endWord, vector<string>& wordList) {

tuple<string, int> node;

unordered_set<string> visited;

queue<tuple<string, int>> w_queue;

w_queue.push(make_tuple(beginWord,1));

while(!w_queue.empty()){
node=w_queue.front();
w_queue.pop();

string word=get<0>(node);
int level=get<1>(node);

if(visited.find(word)==visited.end()){ // visit only if not visited
if(word==endWord) return level; // Found! return

visited.insert(word); // not visied, visit

for(auto &e: wordList){
// if child is valid, push into queue
if(distanceWord(e, word)==1 && visited.find(e)==visited.end()){
w_queue.push(make_tuple(e, level+1));
}
}
}
}
return 0;
}

// return the number of different characters between two strings
int distanceWord(string a, string b){
if(a.size()!=b.size()) return -1;

int diff=0;
for(int i=0; i<a.size(); i++){
if(a[i]!=b[i])
diff++;
}
return diff;
}
};
``````