Word Ladder II LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Link for the ProblemWord Ladder II– LeetCode Problem

Word Ladder II– LeetCode Problem

Problem:

transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> ... -> sk such that:

  • Every adjacent pair of words differs by a single letter.
  • Every si for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList.
  • sk == endWord

Given two words, beginWord and endWord, and a dictionary wordList, return all the shortest transformation sequences from beginWord to endWord, or an empty list if no such sequence exists. Each sequence should be returned as a list of the words [beginWord, s1, s2, ..., sk].

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"]]
Explanation: There are 2 shortest transformation sequences:
"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 there is no valid transformation sequence.

Constraints:

  • 1 <= beginWord.length <= 5
  • endWord.length == beginWord.length
  • 1 <= wordList.length <= 1000
  • wordList[i].length == beginWord.length
  • beginWordendWord, and wordList[i] consist of lowercase English letters.
  • beginWord != endWord
  • All the words in wordList are unique.
Word Ladder II– LeetCode Solutions
Word Ladder II Solution in C++:
class Solution {
 public:
  vector<vector<string>> findLadders(string beginWord, string endWord,
                                     vector<string>& wordList) {
    vector<vector<string>> ans;
    unordered_set<string> wordSet(begin(wordList), end(wordList));
    queue<vector<string>> paths{{{beginWord}}};  // {{"hit"}}

    while (!paths.empty()) {
      unordered_set<string> currentLevelVisited;
      for (int size = paths.size(); size > 0; --size) {
        vector<string> path = paths.front();
        paths.pop();                    // {"hit"}
        string lastWord = path.back();  // "hit"
        for (int i = 0; i < lastWord.length(); ++i) {
          char cache = lastWord[i];  // cache = 'i'
          for (char c = 'a'; c <= 'z'; ++c) {
            lastWord[i] = c;                // "hit" -> "hot" (temporarily)
            if (wordSet.count(lastWord)) {  // find "hot" in wordSet
              currentLevelVisited.insert(lastWord);  // mark "hot" as visited
              vector<string> nextPath(path);
              nextPath.push_back(lastWord);  // nextPath = {"hit", "hot"}
              if (lastWord == endWord)
                ans.push_back(nextPath);
              else
                paths.push(nextPath);
            }
          }
          lastWord[i] = cache;  // "hot" back to "hit"
        }
      }
      for (const string& word : currentLevelVisited)
        wordSet.erase(word);
    }

    return ans;
  }
};
Word Ladder II Solution in Java:
class Solution {
  public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {
    Set<String> wordSet = new HashSet<>(wordList);
    if (!wordSet.contains(endWord))
      return new ArrayList<>();

    Map<String, List<String>> parentToChildren = new HashMap<>();
    Set<String> currentLevelWords = new HashSet<>();
    currentLevelWords.add(beginWord);
    boolean isFound = false;

    while (!currentLevelWords.isEmpty()) {
      // remove words in current level
      for (final String word : currentLevelWords)
        wordSet.remove(word);
      Set<String> nextLevelWords = new HashSet<>();
      // `parent` will be used as a key in `parentToChildren`
      for (final String parent : currentLevelWords) {
        StringBuilder sb = new StringBuilder(parent);
        for (int i = 0; i < sb.length(); ++i) {
          final char cache = sb.charAt(i);
          for (char c = 'a'; c <= 'z'; ++c) {
            sb.setCharAt(i, c);
            final String child = sb.toString();
            if (wordSet.contains(child)) {
              if (child.equals(endWord))
                isFound = true;
              nextLevelWords.add(child);
              parentToChildren.computeIfAbsent(parent, k -> new ArrayList<>()).add(child);
            }
          }
          sb.setCharAt(i, cache);
        }
        currentLevelWords = nextLevelWords;
      }
      if (isFound)
        break;
    }

    if (!isFound)
      return new ArrayList<>();

    List<List<String>> ans = new ArrayList<>();
    List<String> path = new ArrayList<>(Arrays.asList(beginWord));

    dfs(parentToChildren, beginWord, endWord, path, ans);

    return ans;
  }

  // construct the ans by `parentToChildren`
  private void dfs(Map<String, List<String>> parentToChildren, final String word,
                   final String endWord, List<String> path, List<List<String>> ans) {
    if (word.equals(endWord)) {
      ans.add(new ArrayList<>(path));
      return;
    }
    if (!parentToChildren.containsKey(word))
      return;

    for (final String child : parentToChildren.get(word)) {
      path.add(child);
      dfs(parentToChildren, child, endWord, path, ans);
      path.remove(path.size() - 1);
    }
  }
}
Word Ladder II Solution in Python:
class Solution:
  def findLadders(self, beginWord: str, endWord: str, wordList: List[str]) -> List[List[str]]:
    def dfs(word: str, path: List[str]) -> None:
      if word == endWord:
        ans.append(path)
        return
      if word not in dict:
        return

      for child in dict[word]:
        dfs(child, path + [child])

    ans = []
    wordList = set(wordList)

    if endWord not in wordList:
      return ans

    set1 = set([beginWord])
    dict = defaultdict(list)
    isFound = False

    while set1 and not isFound:
      for word in set1:
        wordList.discard(word)
      tempSet = set()
      for parent in set1:
        for i in range(len(parent)):
          for j in string.ascii_lowercase:
            newWord = parent[:i] + j + parent[i + 1:]
            if newWord == endWord:
              dict[parent].append(newWord)
              isFound = True
            elif newWord in wordList and not isFound:
              tempSet.add(newWord)
              dict[parent].append(newWord)
      set1 = tempSet

    if isFound:
      dfs(beginWord, [beginWord])

    return ans

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