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);
              }
          }
      }
  }

Amazon session

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