A 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
sifor1 <= i <= kis inwordList. Note thatbeginWorddoes not need to be inwordList. 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 <= 10endWord.length == beginWord.length1 <= wordList.length <= 5000wordList[i].length == beginWord.lengthbeginWord,endWord, andwordList[i]consist of lowercase English letters.beginWord != endWord- All the words in
wordListare unique.
Solution:
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/**
* The logic is to preprocess the wordList in the form of *
* Eg:WordList: ["hot","dot","dog","lot","log","cog"]
* The Preprocessed output will have the following map. This is nothing but the adjacencyList for the graph.
* {do*=[dot, dog], d*g=[dog], c*g=[cog], ho*=[hot], *og=[dog, log, cog], h*t=[hot],
* lo*=[lot, log], l*t=[lot], l*g=[log], *ot=[hot, dot, lot], d*t=[dot], co*=[cog]}
**/
class Solution {
public int ladderLength(String beginWord, String end, List<String> wordList) {
Map<String, List<String>> map = preProcess(wordList);
System.out.println(map);
//BFS from beginWord.
Set<String> visited = new HashSet<>();
Queue<String[]> q = new LinkedList<>();
q.add(new String[]{beginWord, "0"});
Integer res = null;
while (q.size() != 0) {
String[] node = q.poll();
String cur = node[0];
int level = Integer.parseInt(node[1]);
if (cur.equals(end)) {
res = level;
break;
}
if (visited.contains(cur)) {
continue;
}
visited.add(cur);
for (int i = 0; i < cur.length(); i++) {
String temp = cur.substring(0, i) + "*" + cur.substring(i + 1, cur.length());
List<String> list = map.get(temp);
// System.out.println(cur + ":" + temp + ":" + list);
if (list != null) {
for (String s : list) {
q.add(new String[]{s, (level + 1) + ""});
}
}
}
}
return res == null ? 0 : res + 1;
}
Map<String, List<String>> preProcess(List<String> words) {
Map<String, List<String>> map = new HashMap<>();
for (String w : words) {
for (int i = 0; i < w.length(); i++) {
String temp = w.substring(0, i) + "*" + w.substring(i + 1, w.length());
List<String> list = map.getOrDefault(temp, new ArrayList<>());
list.add(w);
map.put(temp, list);
}
}
return map;
}
}
