Introduction

In this blog post, we will explore how to solve the LeetCode problem "127" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> ... -> sk such that: Every adjacent pair of words differs by a single letter. Every si for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList. sk == endWord Given two words, beginWord and endWord, and a dictionary wordList, return the number of words in the shortest transformation sequence from beginWord to endWord, or 0 if no such sequence exists. Example 1: Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"] Output: 5 Explanation: One shortest transformation sequence is "hit" -> "hot" -> "dot" -> "dog" -> cog", which is 5 words long. Example 2: Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"] Output: 0 Explanation: The endWord "cog" is not in wordList, therefore there is no valid transformation sequence. Constraints: 1 <= beginWord.length <= 10 endWord.length == beginWord.length 1 <= wordList.length <= 5000 wordList[i].length == beginWord.length beginWord, endWord, and wordList[i] consist of lowercase English letters. beginWord != endWord All the words in wordList are unique.

Explanation

Here's a solution to the word ladder problem, focusing on efficiency and clarity:

• Breadth-First Search (BFS): The core idea is to treat the words as nodes in a graph, where an edge exists between two words if they differ by only one letter. BFS is ideal for finding the shortest path in an unweighted graph.

• Optimized Neighbor Generation: Instead of iterating through the entire wordList to find neighbors for each word, we can generate potential neighbors by systematically changing each letter in the current word and checking if the resulting word exists in the wordList. This significantly reduces the search space.

• Runtime Complexity: O(M*N*26), where M is the length of words and N is the number of words in wordList. Storage Complexity: O(N), where N is the number of words in wordList.

Code

    from collections import deque

def ladderLength(beginWord: str, endWord: str, wordList: list[str]) -> int:
    """
    Finds the length of the shortest transformation sequence from beginWord to endWord.
    """

    if endWord not in wordList:
        return 0

    wordList = set(wordList)  # Convert to set for faster lookups
    queue = deque([(beginWord, 1)])  # (word, level)
    visited = {beginWord}

    while queue:
        word, level = queue.popleft()

        if word == endWord:
            return level

        for i in range(len(word)):
            for char_code in range(ord('a'), ord('z') + 1):
                char = chr(char_code)
                neighbor = word[:i] + char + word[i+1:]

                if neighbor in wordList and neighbor not in visited:
                    queue.append((neighbor, level + 1))
                    visited.add(neighbor)

    return 0