Word Ladder

Naive Solution: Brute-Force

The most straightforward approach is to explore all possible transformation sequences from beginWord until we reach endWord. This can be done using a graph-like exploration, where each word in wordList is a node, and edges connect words that differ by only one letter. A brute-force method would involve generating all possible paths and checking their validity.

1. Build the Graph: Construct a graph where nodes are words, and edges connect words with a single-letter difference.
2. Explore All Paths: Use Depth-First Search (DFS) or Breadth-First Search (BFS) to explore all paths from beginWord to endWord.
3. Find the Shortest: Keep track of the shortest valid sequence found so far.

Big(O) Analysis:

• Time Complexity: O(V!), where V is the number of words in wordList. In the worst case, you might explore all possible permutations of words.
• Space Complexity: O(V), to store the graph and the current path.

This method is highly inefficient and impractical for larger dictionaries due to the exponential time complexity.

Optimal Solution: Breadth-First Search (BFS)

A more efficient approach involves using Breadth-First Search (BFS) to find the shortest transformation sequence. BFS guarantees that the first time we encounter endWord, we've found the shortest path.

1. Initialization:
   • Create a queue to store words to explore.
   • Initialize the queue with beginWord and a distance of 1.
   • Use a set to keep track of visited words to avoid cycles.
2. BFS Traversal:
   • While the queue is not empty:
     • Dequeue a word and its distance.
     • For each word in wordList:
       • If the word has not been visited and differs by only one letter from the current word:
         • If the word is endWord, return the current distance + 1.
         • Enqueue the word with the updated distance (distance + 1).
         • Mark the word as visited.
3. No Path Found:
   • If the queue becomes empty and endWord is not found, return 0.

Big(O) Analysis:

• Time Complexity: O(M² * N), where N is the number of words in wordList and M is the length of each word. The M² comes from checking if two words differ by one letter (M to iterate through the letters, and another M in the worst case of string slicing). We do this for all N words in the worst case.
• Space Complexity: O(N), for the queue and the visited set, in the worst case, may contain all the words in wordList.

Edge Cases:

• If endWord is not in wordList, there is no transformation, so return 0.
• If beginWord is equal to endWord, return 0 (as the problem states beginWord != endWord).
• If wordList is empty, and beginWord is not endWord, return 0 (no possible transformations).

Code (Python):

from collections import deque

def ladderLength(beginWord: str, endWord: str, wordList: list[str]) -> int:
    if endWord not in wordList:
        return 0

    wordList = set(wordList)
    queue = deque([(beginWord, 1)])
    visited = {beginWord}

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

        if word == endWord:
            return distance

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

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

    return 0