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 spell checkers, word games, and natural language processing tasks.

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 suitable here because it explores all nodes at the present depth level before moving on to nodes at the next depth level, ensuring the shortest path is found.

Initial naive solution might involve a Depth-First Search (DFS), but it is not optimal due to its potential to explore deeper paths before finding the shortest one.

Using BFS, we can efficiently find the shortest transformation sequence. The BFS approach involves:

- Using a queue to explore each word level by level.
- Tracking visited words to avoid cycles.
- Generating all possible one-letter transformations and checking if they exist in the word list.

Here is a step-by-step breakdown of the BFS algorithm:

- Initialize a queue and add the
*beginWord*with a step count of 1. - Initialize a set to track visited words.
- While the queue is not empty, dequeue the front word and generate all possible one-letter transformations.
- If a transformation matches the
*endWord*, return the current step count + 1. - If the transformation exists in the word list and has not been visited, add it to the queue and mark it as visited.
- If the queue is exhausted without finding the
*endWord*, return 0.

```
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);
Set<String> visited = new HashSet<>();
visited.add(beginWord);
int steps = 1;
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
String currentWord = queue.poll();
char[] currentChars = currentWord.toCharArray();
for (int j = 0; j < currentChars.length; j++) {
char originalChar = currentChars[j];
for (char c = 'a'; c <= 'z'; c++) {
if (c == originalChar) continue;
currentChars[j] = c;
String newWord = new String(currentChars);
if (newWord.equals(endWord)) {
return steps + 1;
}
if (wordSet.contains(newWord) && !visited.contains(newWord)) {
queue.add(newWord);
visited.add(newWord);
}
}
currentChars[j] = originalChar;
}
}
steps++;
}
return 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 total 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 queue and the visited set.

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 involves ensuring the algorithm correctly identifies when no valid transformation exists and handles the constraints appropriately.

To test the solution comprehensively, consider a variety of test cases:

- Simple cases with direct transformations.
- Cases with multiple intermediate transformations.
- Edge cases where no transformation is possible.

Using testing frameworks 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.
- Think about the most efficient way to explore all possible solutions (e.g., BFS for shortest path problems).
- Practice solving similar problems to develop a deeper understanding of common patterns and algorithms.

Understanding and solving the Word Ladder problem involves recognizing the need for an efficient search algorithm like BFS. By breaking down the problem, considering edge cases, and thoroughly testing the solution, you can develop a robust approach to finding the shortest transformation sequence.

Practice and exploration of similar problems will further enhance your problem-solving skills and algorithmic thinking.

For further reading and practice, consider the following resources: