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```package com.thealgorithms.strings;

import java.util.Arrays;
import java.util.HashSet;
import java.util.List;
import java.util.Queue;

/*
**Problem Statement:**
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.
*/

/**
* This function finds the ladderLength
*
* @param beginWord: Starting word of the ladder
* @param endWord: Ending word of the ladder
* @param wordList: This list contains the words which needs to be included
* if the endword is there. Otherwise, will return the length as 0.
*/
public static int ladderLength(String beginWord, String endWord, List<String> wordList) {
HashSet<String> set = new HashSet();
for (String word : wordList) {
}

if (!set.contains(endWord)) {
return 0;
}

queue.offer(beginWord);
int level = 1;

while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
String curr = queue.poll();
char[] words_chars = curr.toCharArray();
for (int j = 0; j < words_chars.length; j++) {
char original_chars = words_chars[j];
for (char c = 'a'; c <= 'z'; c++) {
if (words_chars[j] == c) {
continue;
}
words_chars[j] = c;
String new_word = String.valueOf(words_chars);
if (new_word.equals(endWord)) {
return level + 1;
}
if (set.contains(new_word)) {
set.remove(new_word);
queue.offer(new_word);
}
}
words_chars[j] = original_chars;
}
}
level++;
}
return 0;
}
}
```