Word Ladder - Java Solution and Time Complexity Analysis

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: 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.

Understanding the Problem

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.

Approach

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 in this case, is represented by words and their transformations.

Initial naive solution might involve generating all possible transformations and checking each one, but this is highly inefficient. Instead, we use BFS to explore all possible transformations level by level.

Optimized Solution

We use a queue to perform BFS and a set to keep track of visited words to avoid cycles. For each word, we generate all possible one-letter transformations and check if they are in the word list. If we find the endWord, we return the current transformation length.

Algorithm

Initialize a queue and add the beginWord with a transformation length of 1.
Initialize a set with the word list for quick lookup.
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 length + 1.
If a transformation is in the word list and not visited, add it to the queue and mark it as visited.
If the queue is exhausted without finding the endWord, return 0.

Code Implementation

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 length = 1;

        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                String word = queue.poll();
                char[] wordChars = word.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 length + 1;
                        }
                        if (wordSet.contains(newWord)) {
                            queue.add(newWord);
                            wordSet.remove(newWord);
                        }
                    }
                    wordChars[j] = originalChar;
                }
            }
            length++;
        }
        return 0;
    }
}

Complexity Analysis

The time complexity of this 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 generate M possible transformations, and for each transformation, we check against the word list.

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

Edge Cases

Potential edge cases include:

The endWord not being in the word list.
All words being the same length but no possible transformations.
Empty word list.

Each of these cases is handled by the algorithm, ensuring robustness.

Testing

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

Simple transformations with a small word list.
Large word lists with multiple possible paths.
Edge cases as mentioned above.

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

Thinking and Problem-Solving Tips

When approaching such problems, it's crucial to:

Understand the problem constraints and requirements thoroughly.
Consider different algorithms and their suitability for the problem.
Break down the problem into smaller, manageable parts.
Practice similar problems to improve problem-solving skills.

Conclusion

Understanding and solving the Word Ladder problem is an excellent exercise in graph traversal and BFS. It highlights the importance of choosing the right algorithm for the problem and optimizing for efficiency. Practice and exploration of similar problems can further enhance problem-solving skills.

Additional Resources

For further reading and practice, consider the following resources: