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:**

- 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:5Explanation: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:0Explanation:The endWord "cog" is not in wordList, therefore no possibletransformation.

The core challenge of this problem is to find the shortest path from the *beginWord* to the *endWord* by changing only one letter at a time, with each intermediate word existing in the given word list. This problem is significant in various applications such as word games, natural language processing, and AI-based text generation.

Potential pitfalls include not considering all possible transformations or not efficiently searching through the word list, leading to suboptimal solutions.

To solve this problem, we can use the Breadth-First Search (BFS) algorithm. BFS is ideal for finding the shortest path in an unweighted graph, which aligns with our need to find the shortest transformation sequence.

Here’s a step-by-step approach:

- Use a queue to perform BFS, starting with the
*beginWord*. - Track the level (or depth) of each word transformation.
- For each word, generate all possible one-letter transformations and check if they exist in the word list.
- If a transformation matches the
*endWord*, return the current level + 1. - Continue the process until the queue is empty or the transformation is found.

Here’s a detailed breakdown of the BFS algorithm:

- Initialize a queue and add the
*beginWord*with level 1. - Use a set to store the word list for O(1) lookups.
- While the queue is not empty, dequeue the front element.
- For each character in the word, try changing it to every letter from 'a' to 'z'.
- If the new word is in the word list, add it to the queue and remove it from the word list to prevent reprocessing.
- If the new word matches the
*endWord*, return the current level + 1.

```
import java.util.*;
public class WordLadder {
public int ladderLength(String beginWord, String endWord, List<String> wordList) {
Set<String> wordSet = new HashSet<>(wordList);
if (!wordSet.contains(endWord)) {
return 0;
}
Queue<String> queue = new LinkedList<>();
queue.add(beginWord);
int level = 1;
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
String currentWord = queue.poll();
char[] wordChars = currentWord.toCharArray();
for (int j = 0; j < wordChars.length; j++) {
char originalChar = wordChars[j];
for (char c = 'a'; c <= 'z'; c++) {
if (wordChars[j] == c) continue;
wordChars[j] = c;
String newWord = new String(wordChars);
if (newWord.equals(endWord)) {
return level + 1;
}
if (wordSet.contains(newWord)) {
queue.add(newWord);
wordSet.remove(newWord);
}
}
wordChars[j] = originalChar;
}
}
level++;
}
return 0;
}
public static void main(String[] args) {
WordLadder wl = new WordLadder();
List<String> wordList1 = Arrays.asList("hot", "dot", "dog", "lot", "log", "cog");
System.out.println(wl.ladderLength("hit", "cog", wordList1)); // Output: 5
List<String> wordList2 = Arrays.asList("hot", "dot", "dog", "lot", "log");
System.out.println(wl.ladderLength("hit", "cog", wordList2)); // Output: 0
}
}
```

The time complexity of the BFS approach is O(M^2 * N), where M is the length of each word and N is the number of words in the word list. This is because for each word, we are generating M possible transformations, and for each transformation, we are checking against the word list.

The space complexity is O(M * N) due to the storage of the word list in a set and the queue used for BFS.

Potential edge cases include:

- The
*endWord*not being in the word list, which should return 0. - All words being the same length and containing only lowercase alphabetic characters.
- Handling cases where no transformation sequence is possible.

Testing for these edge cases ensures the robustness of the solution.

To test the solution comprehensively, consider the following test cases:

- Basic cases with a clear transformation sequence.
- Cases where the
*endWord*is not in the word list. - Cases with the minimum and maximum possible word lengths.

Using a testing framework like JUnit can help automate and validate these test cases effectively.

When approaching such problems, consider the following tips:

- Break down the problem into smaller, manageable parts.
- Use diagrams or pseudo-code to visualize the solution.
- Practice similar problems to improve problem-solving skills.

Understanding and solving the Word Ladder problem helps improve skills in graph traversal algorithms like BFS. It also enhances problem-solving abilities and prepares for more complex algorithmic challenges.

Practice and exploration of similar problems are encouraged to gain deeper insights and proficiency.

For further reading and practice, consider the following resources: