Word Ladder III - JavaScript Solution with 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 it is not optimal due to its potential to explore longer paths first, leading to inefficiencies.

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

Algorithm

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

Initialize a queue with the beginWord and a level counter.
Use a set to keep track of visited words to avoid cycles.
While the queue is not empty, dequeue the front element.
For each character in the current word, try changing it to every other possible character from 'a' to 'z'.
If the new word is the endWord, return the current level + 1.
If the new word 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

// JavaScript implementation of the BFS approach
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 with the beginWord and level 1
    const queue = [[beginWord, 1]];
    const visited = new Set();
    visited.add(beginWord);

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

        // Try changing each character 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 endWord, return the current level + 1
                if (newWord === endWord) return level + 1;

                // If the new word is in the wordSet and not visited, add it to the queue
                if (wordSet.has(newWord) && !visited.has(newWord)) {
                    queue.push([newWord, level + 1]);
                    visited.add(newWord);
                }
            }
        }
    }

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

Complexity Analysis

The time complexity of this BFS 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 character transformations. The space complexity is O(N) due to the queue and visited set.

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 algorithm correctly identifies when no valid transformation exists and handles various word lengths and character sets.

Testing

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 Jest or Mocha can help automate and validate these test cases effectively.

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 data structures to use (e.g., sets for quick lookups).
Practice similar problems to develop a deeper understanding of common patterns and algorithms.

Conclusion

Understanding and solving the Word Ladder problem using BFS provides valuable insights into graph traversal algorithms and their applications. Practicing such problems helps improve problem-solving skills and prepares you for more complex challenges.

Additional Resources

For further reading and practice, consider exploring the following resources: