126. Word Ladder II.md

126. Word Ladder II

Question:

Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) 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.

Note:

1. Return an empty list if there is no such transformation sequence.
2. All words have the same length.
3. All words contain only lowercase alphabetic characters.
4. You may assume no duplicates in the word list.
5. You may assume beginWord and endWord are non-empty and are not the same.

Example 1:

Input:
beginWord = "hit",
endWord = "cog",
wordList = ["hot","dot","dog","lot","log","cog"]

Output:
[
  ["hit","hot","dot","dog","cog"],
  ["hit","hot","lot","log","cog"]
]

Example 2:

Input:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

Output: []

Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.

Solutions

• Method 1: dfs TLE

   class Solution {
   	private int min = Integer.MAX_VALUE >> 1;
   	public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {
   		List<List<String>> result = new ArrayList<>();
   		Set<String> words = new HashSet<>(wordList);
   		if(!words.contains(endWord)) return result;
   		int len = beginWord.length();
   		List<String> temp = new ArrayList<String>();
   		temp.add(beginWord);
   		dfs(result, words, len, endWord, beginWord, temp);
   		return result;
   	}
   	private void dfs(List<List<String>> result, Set<String> words, int len, String endWord, String cur, List<String> temp){
   		if(cur.equals(endWord)){
   			if(temp.size() < min){
   				result.clear();
   				min = temp.size();
   			}
   			result.add(new ArrayList<String>(temp));
   		}else if(temp.size() < min){
   			for(int i = 0; i < len; i++){
   				char[] arr = cur.toCharArray();
   				for(char c = 'a'; c <= 'z'; c++){
   					arr[i] = c;
   					String next = new String(arr);
   					if(words.contains(next)){
   						words.remove(next);
   						temp.add(next);
   						dfs(result, words, len, endWord, next, temp);
   						temp.remove(temp.size() - 1);
   						words.add(next);
   					}
   				}
   				arr[i] = cur.charAt(i);
   			}
   		}
   	}
   }

• Method 2: bfs AC

   class Solution {
   	private Map<String, List<String>> map;
   	public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {
   		List<List<String>> result = new LinkedList<>();
   		Set<String> words = new HashSet<>(wordList);
   		if(!words.contains(endWord)) return result;
   		words.remove(beginWord);
   		LinkedList<String> queue = new LinkedList<>();
   		map = new HashMap<>();
   		int len = beginWord.length();
   		boolean found = false;
   		queue.add(beginWord);
   		while(!queue.isEmpty() && !found){
   			int size = queue.size();
   			Set<String> curLevel = new HashSet<>();
   			for(int i = 0; i < size; i++){
   				String cur = queue.poll();
   				if(cur.equals(endWord)) found = true;   
   				char[] arr = cur.toCharArray();
   				for(int l = 0; l < len; l++){
   					for(char c = 'a'; c <= 'z'; c++){
   						if(c == cur.charAt(l)) continue;
   						arr[l] = c;
   						String next = new String(arr);
   						if(words.contains(next) || curLevel.contains(next)){
   							map.put(cur, !map.containsKey(cur) ? new ArrayList<String>(): map.get(cur));
   							map.get(cur).add(next);
   							curLevel.add(next);
   							if(words.contains(next)) queue.offer(next);
   							words.remove(next);
   						}
   					}
   					arr[l] = cur.charAt(l);
   				}
   			}
   		}
   		List<String> temp = new LinkedList<>();
   		temp.add(beginWord);
   		dfs(result, temp, endWord, beginWord);
   		return result;
   	}
   	private void dfs(List<List<String>> result, List<String> temp, String endWord, String cur){
   		if(cur.equals(endWord)){
   			result.add(new ArrayList<>(temp));
   		}else{
   			if(!map.containsKey(cur)) return;
   			List<String> adj = map.get(cur);
   			for(String s: adj){
   				temp.add(s);
   				dfs(result, temp, endWord, s);
   				temp.remove(temp.size() - 1);
   			}
   		}
   	}
   }

Second Time

• Method 1: bfs + dfs

  class Solution {
      private List<List<String>> result;
      private Map<String, List<String>> map;
      public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {
          Set<String> wordSet = new HashSet<>(wordList);
          wordSet.remove(beginWord);
          this.result = new ArrayList<>();
          if(!wordSet.contains(endWord)) return this.result;
          this.map = new HashMap<>();
          Queue<String> q = new LinkedList<>();
          q.offer(beginWord);
          boolean found = false;
          int len = beginWord.length();
          while(!found && !q.isEmpty()){
              int size = q.size();
              Set<String> nextLevel = new HashSet<>();
              for(int i = 0; i < size; i++){
                  String temp = q.poll();
                  if(temp.equals(endWord)) found = true;
                  char[] arr = temp.toCharArray();
                  for(int j = 0; j < len; j++){
                      char origin = arr[j];
                      for(char c = 'a'; c <= 'z'; c++){
                          if(c == origin) continue;
                          arr[j] = c;
                          String next = new String(arr);
                          if(wordSet.contains(next) || nextLevel.contains(next)){
                              List<String> list = map.containsKey(temp) ? map.get(temp): new ArrayList<>();
                              list.add(next);
                              map.put(temp, list);
                              nextLevel.add(next);
                              if(wordSet.contains(next)) q.offer(next);
                              wordSet.remove(next);
                          }
                      }
                      arr[j] = origin;
                  }
              }
          }
          List<String> temp = new ArrayList<>();
          temp.add(beginWord);
          dfs(temp, endWord, beginWord);
          return this.result;
      }
      private void dfs(List<String> temp, String end, String cur){
          if(cur.equals(end)) this.result.add(new ArrayList<String>(temp));
          else{
              if(!map.containsKey(cur)) return;
              List<String> adj = map.get(cur);
              for(String s: adj){
                  temp.add(s);
                  dfs(temp, end, s);
                  temp.remove(temp.size() - 1);
              }
          }
      }
  }