Word Ladder II - Python 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 Breadth-First Search (BFS), which is well-suited for finding the shortest path in an unweighted graph. Here, each word represents a node, and an edge exists between two nodes if they differ by exactly one letter.

Naive Solution

A naive solution would involve generating all possible transformations for each word and checking if they exist in the word list. This approach is not optimal due to its high time complexity.

Optimized Solution

We can optimize the solution using BFS. The BFS will explore all possible transformations level by level, ensuring that the first time we reach the endWord, we have found the shortest path.

Algorithm

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

Initialize a queue with the beginWord and a set for the word list.
While the queue is not empty, dequeue the front element.
For each character in the current word, try changing it to every other character from 'a' to 'z'.
If the new word is in the word list, add it to the queue and remove it from the word list to prevent revisiting.
If the new word is the endWord, return the current level + 1.
If the queue is exhausted without finding the endWord, return 0.

Code Implementation

import collections

def ladderLength(beginWord, endWord, wordList):
    # Convert wordList to a set for O(1) lookups
    wordSet = set(wordList)
    if endWord not in wordSet:
        return 0
    
    # Initialize the queue for BFS
    queue = collections.deque([(beginWord, 1)])
    
    while queue:
        current_word, level = queue.popleft()
        
        # Try changing each character of the current word
        for i in range(len(current_word)):
            for c in 'abcdefghijklmnopqrstuvwxyz':
                next_word = current_word[:i] + c + current_word[i+1:]
                
                # If the next word is the end word, return the level + 1
                if next_word == endWord:
                    return level + 1
                
                # If the next word is in the word set, add it to the queue and remove from the set
                if next_word in wordSet:
                    wordSet.remove(next_word)
                    queue.append((next_word, level + 1))
    
    return 0

Complexity Analysis

The time complexity of this BFS 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 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 queue and the set used to store the word list.

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.

These cases are handled by the initial checks and the BFS logic.

Testing

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

Simple cases with a direct transformation.
Cases where the endWord is not in the word list.
Cases with multiple possible paths to the endWord.

Using a testing framework like unittest in Python can help automate and validate these test cases.

Thinking and Problem-Solving Tips

When approaching such problems, it is crucial to:

Understand the problem constraints and requirements.
Consider different algorithms and their suitability for the problem.
Break down the problem into smaller, manageable parts.
Write and test code incrementally to catch errors early.

Practicing similar problems and studying various algorithms can significantly improve problem-solving skills.

Conclusion

In this blog post, we discussed the Word Ladder II problem, explored different approaches to solve it, and provided a detailed BFS-based solution in Python. Understanding and solving such problems is essential for developing strong algorithmic thinking and coding skills.

Additional Resources

For further reading and practice, consider the following resources: