LeetCode Solutions
127. Word Ladder
Time: $O(|\texttt{wordList}| \cdot 26^{\texttt{wordList[i]}})$ Space: $O(|\texttt{wordList}|)$
class Solution {
public:
int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
unordered_set<string> wordSet(begin(wordList), end(wordList));
if (!wordSet.count(endWord))
return 0;
int ans = 0;
queue<string> q{{beginWord}};
while (!q.empty()) {
++ans;
for (int sz = q.size(); sz > 0; --sz) {
string word = q.front();
q.pop();
for (int i = 0; i < word.length(); ++i) {
const char cache = word[i];
for (char c = 'a'; c <= 'z'; ++c) {
word[i] = c;
if (word == endWord)
return ans + 1;
if (wordSet.count(word)) {
q.push(word);
wordSet.erase(word);
}
}
word[i] = cache;
}
}
}
return 0;
}
};
class Solution {
public int ladderLength(String beginWord, String endWord, List<String> wordList) {
Set<String> wordSet = new HashSet<>(wordList);
if (!wordSet.contains(endWord))
return 0;
int ans = 0;
Queue<String> q = new ArrayDeque<>(Arrays.asList(beginWord));
while (!q.isEmpty()) {
++ans;
for (int sz = q.size(); sz > 0; --sz) {
StringBuilder sb = new StringBuilder(q.poll());
for (int i = 0; i < sb.length(); ++i) {
final char cache = sb.charAt(i);
for (char c = 'a'; c <= 'z'; ++c) {
sb.setCharAt(i, c);
final String word = sb.toString();
if (word.equals(endWord))
return ans + 1;
if (wordSet.contains(word)) {
q.offer(word);
wordSet.remove(word);
}
}
sb.setCharAt(i, cache);
}
}
}
return 0;
}
}
class Solution:
def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
wordSet = set(wordList)
if endWord not in wordSet:
return 0
ans = 0
q = deque([beginWord])
while q:
ans += 1
for _ in range(len(q)):
wordList = list(q.popleft())
for i, cache in enumerate(wordList):
for c in string.ascii_lowercase:
wordList[i] = c
word = ''.join(wordList)
if word == endWord:
return ans + 1
if word in wordSet:
q.append(word)
wordSet.remove(word)
wordList[i] = cache
return 0