Word Ladder IV - 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.
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 depth + 1.
If the transformation exists 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’s a detailed breakdown of the BFS algorithm:

Initialize the queue with the beginWord and a depth of 1.
While the queue is not empty:
1. Dequeue the front element (current word and depth).
2. For each character in the current word, try changing it to every letter from 'a' to 'z'.
3. If the new word matches the endWord, return the current depth + 1.
4. If the new word exists 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 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 storage of the word list in a set and the queue used for BFS.

Edge Cases

Potential edge cases include:

The endWord is not in the word list. The function should return 0.
The beginWord is the same as the endWord. The function should return 1.
All words are of different lengths. This violates the problem constraints and should not occur.

Testing

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

def test_ladderLength():
    assert ladderLength("hit", "cog", ["hot","dot","dog","lot","log","cog"]) == 5
    assert ladderLength("hit", "cog", ["hot","dot","dog","lot","log"]) == 0
    assert ladderLength("a", "c", ["a","b","c"]) == 2
    assert ladderLength("hit", "hit", ["hit"]) == 1
    print("All test cases pass")

test_ladderLength()

Thinking and Problem-Solving Tips

When approaching such problems, consider the following tips:

Break down the problem into smaller, manageable parts.
Use appropriate data structures for efficient lookups and modifications.
Think about edge cases and how to handle them.
Practice similar problems to improve problem-solving skills.

Conclusion

Understanding and solving the Word Ladder problem helps in developing skills for graph traversal and shortest path algorithms. It is a common problem in technical interviews and has practical applications in various fields.

Practice and explore further to deepen your understanding and improve your problem-solving abilities.