[LeetCode] Word Ladder

Problem:

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.

Solution:O(n)

public class Solution {
   public int ladderLength(String start, String end, HashSet<String> dict) {
      if (dict.size() == 0)  
         return 0; 
  
      LinkedList<String> wordQueue = new LinkedList<>();
      LinkedList<Integer> distanceQueue = new LinkedList<>();
      wordQueue.add(start);
      distanceQueue.add(1);
  
      while(!wordQueue.isEmpty()){
         String currWord = wordQueue.pop();
  Integer currDistance = distanceQueue.pop();
  
  if(currWord.equals(end)){
     return currDistance;
  }
  
  for(int i=0; i<currWord.length(); i++){
     char[] currCharArr = currWord.toCharArray();
     for(char c='a'; c<='z'; c++){
        currCharArr[i] = c;
        String newWord = new String(currCharArr);
        if(dict.contains(newWord)){
           wordQueue.add(newWord);
           distanceQueue.add(currDistance+1);
           dict.remove(newWord);
        }
     }
  }
      }
      return 0;
   }
}

Key Concept: Like BFS, and use two queue to help tracing