WordLadder.java

package breadth_first_search;

import java.util.*;

/**
 * Created by gouthamvidyapradhan on 21/03/2017.
 Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that:

 Only one letter can be changed at a time.
 Each transformed word must exist in the word list. Note that beginWord is not a transformed word.
 For example,

 Given:
 beginWord = "hit"
 endWord = "cog"
 wordList = ["hot","dot","dog","lot","log","cog"]
 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.
 You may assume no duplicates in the word list.
 You may assume beginWord and endWord are non-empty and are not the same.
 */
public class WordLadder
{

    class State
    {
        String word;
        int len;
        State(String word, int len)
        {
            this.word = word;
            this.len = len;
        }
    }

    private static Queue<State> queue = new ArrayDeque<>();
    private static Set<String> dictionary = new HashSet<>();
    private static final String CONST = "abcdefghijklmnopqrstuvwxyz";
    private static Set<String> done = new HashSet<>();
    /**
     * Main method
     * @param args
     * @throws Exception
     */
    public static void main(String[] args) throws Exception
    {
        List<String> list = new ArrayList<>();
        list.add("hot");
        list.add("dot");
        list.add("dog");
        list.add("lot");
        list.add("log");
        list.add("cog");
        System.out.println(new WordLadder().ladderLength("hit", "cog", list));
    }


    public int ladderLength(String beginWord, String endWord, List<String> wordList)
    {
        dictionary.addAll(wordList);
        queue.offer(new State(beginWord, 0));
        done.add(beginWord);
        while(!queue.isEmpty())
        {
            State head = queue.poll();
            if(head.word.equals(endWord))
                return head.len + 1;
            for(int i = 0, l = CONST.length(); i < l; i ++ )
            {
                StringBuilder word = new StringBuilder(head.word);
                for(int j = 0, ln = word.length(); j < ln; j ++ )
                {
                    char old = word.charAt(j);
                    word.replace(j, j + 1, String.valueOf(CONST.charAt(i)));
                    if(!done.contains(word.toString()))
                    {
                        if(dictionary.contains(word.toString()))
                        {
                            done.add(word.toString());
                            queue.offer(new State(word.toString(), head.len + 1));
                        }
                    }
                    word.replace(j, j + 1, String.valueOf(old));
                }
            }
        }
        return 0;
    }

}