LeetCode 127: Word Ladder

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:

1. Only one letter can be changed at a time
2. Each intermediate word must exist in the dictionary

For example,
Given:
start = "hit"
end = "cog"
dict = ["hot","dot","dog","lot","log"]

As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.
Note:

• Return 0 if there is no such transformation sequence.
• All words have the same length.
• All words contain only lowercase alphabetic characters.

public class Solution {
    public int ladderLength(String start, String end, Set<String> dict) {
        // Adopt breath fisrt search
        return bfs(start, end, dict);
    }
    
    private int bfs(String start, String end, Set<String> dict)
    {
        Queue<String> word = new LinkedList<>();
        Queue<Integer> length = new LinkedList<>();
        
        word.offer(start);
        length.offer(2);
        
        while (!word.isEmpty())
        {
            String v = word.poll();
            int len = length.poll();

            for (int i = 0; i < v.length(); i++)
            {
                // Note: Adopt character replacement strategy can save much time.
                for (char c = 'a'; c <= 'z'; c++)
                {
                    if (v.charAt(i) == c)
                        continue;
                        
                    String w = replace(v, i, c);
                    
                    if (w.equals(end))
                        return len;
                    
                    if (dict.contains(w))
                    {
                        word.offer(w);
                        length.offer(len+1);
                        dict.remove(w);
                    }
                }
            }
        }
        
        return 0;
    } 
    
	private String replace(String s, int index, char c)
	{
        char[] chars = s.toCharArray();
        chars[index] = c;
        return new String(chars);
	}
}