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.
```

## Solutions

``````class Solution {

// Do not pairwise calculate the edit distance of words present in wordList, there is a good change to

public int ladderLength(String beginWord, String endWord, List<String> wordList) {
if (wordList == null || wordList.size() == 0) {
return 0;
}

if (beginWord.equals(endWord)) {
return 1;
}

// FIXME Convert List to Set, otherwise time limit exceed occurs.
Set<String> dict = new HashSet<>(wordList);

Set<String> visited = new HashSet<>();

// Initialize the startup state
queue.offer(beginWord);

int ans = 1;

while (!queue.isEmpty()) {

ans++;

int size = queue.size();
for (int i = 0; i < size; i++) {

String curr = queue.poll();

// BFS all the one step transformed words
for (String word : adjacentWords) {

// exit of the loop
if (word.equals(endWord)) {
return ans;
}

}
}
}

// Once reach here, no such possible transformation sequence is found.

return 0;
}

private List<String> findAdjacent(String word, Set<String> dict, Set<String> visited) {

// Sort out all possible transformations of the given word by change on letter.
// Any new words not visited yet but present in the word dictionary will be adopted.

for (int i = 0; i < word.length(); i++) {
for (char j = 'a'; j <= 'z'; j++) {
if (j == word.charAt(i)) {
continue;
}

String newWord = transform(word, i, j);
if (visited.contains(newWord)) {
continue;
}

if (dict.contains(newWord)) {
}
}
}

}

private String transform(String word, int idx, Character rc) {
String ans = "";
for (int i = 0; i < word.length(); i++) {
ans += (i == idx) ? rc : word.charAt(i);
}

return ans;
}
}
``````