LeetCode //C++ - 126. Word Ladder II

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126. Word Ladder II

A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> $s_1$ -> $s_2$ -> … -> $s_k$ such that:

• Every adjacent pair of words differs by a single letter.
• Every $s_i$ 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, $s_1$, $s_2$, …, $s_k$].
 

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 <= 500
• wordList[i].length == beginWord.length
• beginWord, endWord, and wordList[i] consist of lowercase English letters.
• beginWord != endWord
• All the words in wordList are unique.
• The sum of all shortest transformation sequences does not exceed $10^5$.

From: LeetCode
Link: 126. Word Ladder II

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Solution:

Ideas:

• Initialization: An unordered set is used to quickly check if a word is in the dictionary (dict). A queue (q) manages the BFS starting with beginWord.
• BFS Loop: The BFS iteratively transforms each character of the current word to explore all possible one-letter transformations. If these transformed words are in dict, they’re added to the queue for further exploration. parents keeps track of each word’s predecessors in the path.
• Backtracking Function: Once the shortest paths are identified (found is true), the buildPaths function reconstructs the paths from endWord to beginWord using the parents map and accumulates them in ladders.
• Handling Level-wise Exploration: visitedThisLevel prevents revisiting words in the same BFS level, optimizing the search and avoiding redundant paths.

Code:

class Solution {
public:vector<vector<string>> findLadders(string beginWord, string endWord, vector<string>& wordList) {unordered_set<string> dict(wordList.begin(), wordList.end());vector<vector<string>> ladders;unordered_map<string, vector<string>> parents;queue<string> q;q.push(beginWord);bool found = false;while (!q.empty() && !found) {int levelSize = q.size();unordered_set<string> visitedThisLevel; // To handle duplicates in the current BFS levelfor (int i = 0; i < levelSize; ++i) {string currentWord = q.front();q.pop();string originalWord = currentWord; // Save the original word// Transform each character of the word to find all possible transformationsfor (char& ch : currentWord) {char originalChar = ch; // Save the original characterfor (ch = 'a'; ch <= 'z'; ++ch) {if (ch == originalChar) continue;if (dict.find(currentWord) != dict.end()) {if (currentWord == endWord) {found = true;}if (!found && visitedThisLevel.insert(currentWord).second) {q.push(currentWord);}parents[currentWord].push_back(originalWord);}}ch = originalChar; // Restore the original character}}// Remove visited words of this level from the dictionary to prevent going backwards in BFSfor (const string& word : visitedThisLevel) {dict.erase(word);}}if (found) {vector<string> path;buildPaths(endWord, beginWord, parents, path, ladders);}return ladders;}private:void buildPaths(const string& word, const string& beginWord,unordered_map<string, vector<string>>& parents,vector<string>& path, vector<vector<string>>& ladders) {if (word == beginWord) {path.push_back(word);vector<string> currentPath = path;reverse(currentPath.begin(), currentPath.end());ladders.push_back(currentPath);path.pop_back();return;}path.push_back(word);for (const string& parent : parents[word]) {buildPaths(parent, beginWord, parents, path, ladders);}path.pop_back();}
};

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