Word Ladder - 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.

Here’s a step-by-step approach:

Use a queue to perform BFS. Initialize the queue with the beginWord and a level counter set to 1.
Use a set to keep track of words that have been visited to avoid cycles.
For each word in the queue, generate all possible transformations by changing one letter at a time.
If a transformation matches the endWord, return the current level + 1.
If the transformation exists in the word list and hasn’t been visited, add it to the queue and mark it as visited.
If the queue is exhausted without finding the endWord, return 0.

Algorithm

Here’s a detailed breakdown of the BFS algorithm:

Initialize a queue with the beginWord and a level counter set to 1.
Initialize a set to keep track of visited words.
While the queue is not empty:
1. Dequeue the front word and its level.
2. For each character in the word, generate all possible transformations by changing it to every letter from 'a' to 'z'.
3. If a transformation matches the endWord, return the current level + 1.
4. If the transformation exists in the word list and hasn’t been visited, add it to the queue and mark it as visited.
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]];
    const visited = new Set();
    visited.add(beginWord);

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

        // Generate all possible transformations
        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 and not visited
                if (wordSet.has(newWord) && !visited.has(newWord)) {
                    queue.push([newWord, level + 1]);
                    visited.add(newWord);
                }
            }
        }
    }

    // If no transformation sequence is found
    return 0;
}

Complexity Analysis

The time complexity of this approach is O(N * M * 26), where N is the number of words in the word list, M is the length of each word, and 26 represents the number of possible transformations for each character. The space complexity is O(N) due to the queue and visited set.

Edge Cases

Potential edge cases include:

The endWord is not in the word list.
The beginWord is the same as the endWord.
All words are of length 1.

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

Testing

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 multiple possible transformation sequences.

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.
Use diagrams or pseudo-code to visualize the solution.
Practice similar problems to build familiarity with common patterns and algorithms.

Conclusion

Understanding and solving the Word Ladder problem helps improve problem-solving skills and familiarity with graph traversal algorithms like BFS. Practice and exploration of similar problems can further enhance these skills.

Additional Resources

For further reading and practice, consider the following resources: