Dictionary Word Transformation Path

Problem Overview

You are given:

• a start word
• a target word
• a dictionary of allowed words

Determine whether the words can be connected by a valid transformation sequence. Each step in the sequence must move from one word to another word in the dictionary by changing exactly one letter at a time.

Constraints

• The length of start and target is between 1 and 10,000.
• The dictionary contains between 1 and 10,000 words.
• All strings consist of lowercase English letters.

Examples

Example 1:

Input:

• start = "hit"
• target = "cog"
• dictionary = ["hot","dot","dog","lot","log","cog"]

Output:

• True

Explanation:

• One possible transformation sequence is "hit" -> "hot" -> "dot" -> "dog" -> "cog".

Example 2:

Input:

• start = "hit"
• target = "cog"
• dictionary = ["hot","dot","dog","lot","log"]

Output:

• False

Explanation:

• There is no possible transformation sequence from "hit" to "cog".

Hints

1. Use a breadth-first search (BFS) to explore all possible transformations.
2. Keep track of visited words to avoid revisiting them.
3. Check if the target word can be reached from the start word.

Solution

To solve this problem, you can use a breadth-first search (BFS) algorithm. Here's a high-level approach:

1. Initialize a queue with the start word and a set to keep track of visited words.
2. While the queue is not empty, do the following:
   • Dequeue a word from the queue.
   • If this word is the target word, return True.
   • For each word in the dictionary that differs from the dequeued word by exactly one letter:
     • If it has not been visited, mark it as visited and enqueue it.
3. If you exhaust all possibilities without reaching the target word, return False.

Here's a sample Python code snippet to illustrate the approach:

`python from collections import deque

def canTransform(start, target, dictionary): def neighbors(word): for i in range(len(word)): for c in 'abcdefghijklmnopqrstuvwxyz': neighbor = word[:i] + c + word[i+1:] if neighbor in dictionary and neighbor != word: yield neighbor

queue = deque([start])
visited = set([start])
while queue:
    current = queue.popleft()
    if current == target:
        return True
    for neighbor in neighbors(current):
        if neighbor not in visited:
            visited.add(neighbor)
            queue.append(neighbor)
return False

`

This code defines a helper function neighbors to find all words in the dictionary that differ by exactly one letter from a given word. The main function uses a BFS approach to explore all possible transformations starting from the start word. If the target word is reached, it returns True; otherwise, it returns False after exploring all possibilities.