# Solving Leetcode Interviews in Seconds with AI: Word Ladder


	# 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](https://www.chatmagic.app), 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
	```python
	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
	```