Word Ladder IV - JavaScript 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 aligns with our need to find the shortest transformation sequence.

Initial naive solution might involve a depth-first search (DFS), but this would not guarantee the shortest path and could be inefficient due to the large number of possible transformations.

Optimized solutions involve using BFS to explore all possible transformations level by level, ensuring the shortest path is found.

Algorithm

1. Add the beginWord to a queue and start the BFS process.

2. For each word in the queue, generate all possible transformations by changing one letter at a time.

3. If a transformation matches the endWord, return the current transformation length.

4. If not, add the transformation to the queue and continue the process.

5. If the queue is exhausted without finding the endWord, return 0.

Code Implementation

/**
 * @param {string} beginWord
 * @param {string} endWord
 * @param {string[]} wordList
 * @return {number}
 */
function ladderLength(beginWord, endWord, wordList) {
    // Convert wordList to a set for O(1) lookups
    const wordSet = new Set(wordList);
    if (!wordSet.has(endWord)) return 0;

    // Initialize the queue for BFS
    const queue = [[beginWord, 1]];

    while (queue.length > 0) {
        const [currentWord, level] = queue.shift();

        // Try changing each letter of the current word
        for (let i = 0; i < currentWord.length; i++) {
            for (let c = 97; c <= 122; c++) { // ASCII 'a' to 'z'
                const newWord = currentWord.slice(0, i) + String.fromCharCode(c) + currentWord.slice(i + 1);

                // If the new word is the end word, return the level + 1
                if (newWord === endWord) return level + 1;

                // If the new word is in the word set, add it to the queue and remove from the set
                if (wordSet.has(newWord)) {
                    queue.push([newWord, level + 1]);
                    wordSet.delete(newWord);
                }
            }
        }
    }

    // If we exhaust the queue without finding the end word, return 0
    return 0;
}

Complexity Analysis

The time complexity of this approach is O(N * M^2), where N is the number of words in the word list and M is the length of each word. This is because for each word, we generate M possible transformations, and each transformation requires O(M) time to create and check.

The space complexity is O(N) due to the queue and the set used to store the word list.

Edge Cases

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 function returns the correct output for various inputs, including those with no possible transformations.

Testing

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

Simple transformations with a clear path.
Cases where the endWord is not in the word list.
Large word lists to test performance and efficiency.

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

Thinking and Problem-Solving Tips

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.
Practice similar problems to improve problem-solving skills and familiarity with algorithms like BFS.

Conclusion

Understanding and solving the Word Ladder problem involves using BFS to find the shortest transformation sequence. By breaking down the problem, considering edge cases, and testing thoroughly, you can develop an efficient and effective solution.

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

Additional Resources

For further reading and practice, consider the following resources: