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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.``````

Constraints:

• `1 <= beginWord.length <= 10`
• `endWord.length == beginWord.length`
• `1 <= wordList.length <= 5000`
• `wordList[i].length == beginWord.length`
• `beginWord``endWord`, 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:
wordList.remove(next_word)
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;
len++;
}

return 0;
}
}
``````

### Word Ladder LeetCode Solution in C++

``````class Solution {
public:
int ladderLength(string beginWord, string endWord, vector<string>& wordList)
{
if(find(wordList.begin(),wordList.end(),endWord)==wordList.end())
return 0;
set<string> s;
for(auto i:wordList)
s.insert(i);
queue<string> q;
q.push(beginWord);
int d=0;
while(!q.empty())
{
d++;
int n=q.size();
while(n--)
{
string curr=q.front();
q.pop();
for(int i=0;i<curr.length();i++)
{
string tmp=curr;
for(char c='a';c<='z';c++)
{
tmp[i]=c;
if(tmp==curr)
continue;
if(tmp==endWord)
return d+1;
if(s.find(tmp)!=s.end())
{
q.push(tmp);
s.erase(tmp);
}
}
}
}
}
return 0;
}
};
``````
##### Word Ladder LeetCode Solution Review:

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