127. Word Ladder
Rachit Keshari
2 min read
Que: A transformation sequence from word beginWord
to word endWord
using a dictionary wordList
is a sequence of words beginWord -> s<sub>1</sub> -> s<sub>2</sub> -> ... -> s<sub>k</sub>
such that:
Every adjacent pair of words differs by a single letter.
Every
s<sub>i</sub>
for1 <= i <= k
is inwordList
. Note thatbeginWord
does not need to be inwordList
.s<sub>k</sub> == 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.
Solution:
class Solution {
class Pair{
String str;
int val;
Pair(String _str,int _val){
this.str = _str;
this.val = _val;
}
}
public int ladderLength(String beginWord, String endWord, List<String> wordList) {
Set<String> st = new HashSet<String>();
for(String str: wordList){ st.add(str); }
Queue<Pair> q = new LinkedList<>();
q.offer(new Pair(beginWord,1));
while(!q.isEmpty()){
String str = q.peek().str;
int val = q.peek().val;
q.poll();
if(str.equals(endWord)){ return val; }
for(int i=0;i<str.length();i++){
char[] word = str.toCharArray();
for(char ch='a';ch<='z';ch++){
word[i]=ch;
String tmp = new String(word);
if(st.contains(tmp)){
q.offer(new Pair(tmp,val+1));
st.remove(tmp);
}
}
}
}
return 0;
}
}
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