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Word Ladder LeetCode Solution – Queslers

Problem – Word Ladder

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 the number of words in the shortest transformation sequence from beginWord to endWord, or 0 if no such sequence exists.

Example 1:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
Output: 5
Explanation: One shortest transformation sequence is "hit" -> "hot" -> "dot" -> "dog" -> cog", which is 5 words long.

Example 2:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
Output: 0
Explanation: The endWord "cog" is not in wordList, therefore there is no valid transformation sequence.


  • 1 <= beginWord.length <= 10
  • endWord.length == beginWord.length
  • 1 <= wordList.length <= 5000
  • 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 LeetCode Solution in Python

class Solution(object):
    def ladderLength(self, beginWord, endWord, wordList):
        wordList = set(wordList)
        queue = collections.deque([[beginWord, 1]])
        while queue:
            word, length = queue.popleft()
            if word == endWord:
                return length
            for i in range(len(word)):
                for c in 'abcdefghijklmnopqrstuvwxyz':
                    next_word = word[:i] + c + word[i+1:]
                    if next_word in wordList:
                        queue.append([next_word, length + 1])
        return 0

Word Ladder LeetCode Solution in Java

public class Solution {

public int ladderLength(String beginWord, String endWord, Set<String> wordList) {
	Set<String> beginSet = new HashSet<String>(), endSet = new HashSet<String>();

	int len = 1;
	int strLen = beginWord.length();
	HashSet<String> visited = new HashSet<String>();
	while (!beginSet.isEmpty() && !endSet.isEmpty()) {
		if (beginSet.size() > endSet.size()) {
			Set<String> set = beginSet;
			beginSet = endSet;
			endSet = set;

		Set<String> temp = new HashSet<String>();
		for (String word : beginSet) {
			char[] chs = word.toCharArray();

			for (int i = 0; i < chs.length; i++) {
				for (char c = 'a'; c <= 'z'; c++) {
					char old = chs[i];
					chs[i] = c;
					String target = String.valueOf(chs);

					if (endSet.contains(target)) {
						return len + 1;

					if (!visited.contains(target) && wordList.contains(target)) {
					chs[i] = old;

		beginSet = temp;
	return 0;

Word Ladder LeetCode Solution in C++

class Solution {
    int ladderLength(string beginWord, string endWord, vector<string>& wordList)
            return 0;
        set<string> s;
        for(auto i:wordList)
        queue<string> q;
        int d=0;
            int n=q.size();
                string curr=q.front();
                for(int i=0;i<curr.length();i++)
                    string tmp=curr;
                    for(char c='a';c<='z';c++)
                            return d+1;
        return 0;
Word Ladder LeetCode Solution Review:

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