Word Ladder III - 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 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 depth of 1.
Use a set to store the word list for O(1) lookups.
While the queue is not empty, dequeue the front element.
For each character in the dequeued word, try changing it to every other character from 'a' to 'z'.
If the new word is the endWord, return the current depth + 1.
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 queue is exhausted without finding the endWord, return 0.

Algorithm

Here is a detailed breakdown of the BFS algorithm:

Initialize the queue with the beginWord and a depth of 1.
Convert the word list to a set for efficient lookups.
While the queue is not empty:
1. Dequeue the front element (current word and depth).
2. For each character in the current word:
  1. Try changing it to every other character from 'a' to 'z'.
  2. If the new word is the endWord, return the current depth + 1.
  3. If the new word is in the word list, add it to the queue and remove it from the word list.
If the queue is exhausted without finding the endWord, return 0.

Code Implementation

from collections import deque

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 with the beginWord and depth 1
    queue = deque([(beginWord, 1)])
    
    while queue:
        current_word, depth = 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 endWord, return the depth + 1
                if next_word == endWord:
                    return depth + 1
                
                # If the next word is in the wordSet, add it to the queue and remove from the set
                if next_word in wordSet:
                    wordSet.remove(next_word)
                    queue.append((next_word, depth + 1))
    
    # If the queue is exhausted without finding the endWord, return 0
    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 are trying M possible transformations, and for each transformation, we are iterating through the word list.

The space complexity is O(M * N) due to the storage of the queue and the word set.

Edge Cases

Potential edge cases include:

The endWord is not in the word list. In this case, the function should return 0.
The beginWord and endWord are the same. This is not allowed by the problem constraints.
All words are of the same length and contain only lowercase alphabetic characters.

Testing

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

Basic test case with a valid transformation sequence.
Test case where the endWord is not in the word list.
Test case with the minimum input size.
Test case with a large word list to test performance.

Thinking and Problem-Solving Tips

When approaching such problems, it is crucial to:

Understand the problem constraints and requirements thoroughly.
Consider different algorithms and data structures that can efficiently solve the problem.
Break down the problem into smaller, manageable parts.
Write pseudocode to outline the solution before implementing it.
Test the solution with various test cases to ensure its correctness and efficiency.

Conclusion

In this blog post, we discussed the Word Ladder III problem and provided a detailed solution using the BFS algorithm. We analyzed the time and space complexity, discussed potential edge cases, and provided tips for approaching similar problems. Understanding and solving such problems is crucial for developing strong problem-solving skills and improving algorithmic thinking.

Additional Resources

For further reading and practice, consider the following resources: