There is a new alien language that uses the English alphabet. However, the order among the letters is unknown to you.
You are given a list of strings words from the alien language's dictionary, where the strings in words are sorted lexicographically by the rules of this new language.
Return a string of the unique letters in the new alien language sorted in lexicographically increasing order by the new language's rules. If there is no solution, return "". If there are multiple solutions, return any of them.
Example 1:
Input: words = ["wrt","wrf","er","ett","rftt"] Output: "wertf"
Example 2:
Input: words = ["z","x"] Output: "zx"
Example 3:
Input: words = ["z","x","z"]
Output: ""
Explanation: The order is invalid, so return "".
Constraints:
1 <= words.length <= 1001 <= words[i].length <= 100words[i]consists of only lowercase English letters.
class Solution:
def alienOrder(self, words: List[str]) -> str:
g = [[False] * 26 for _ in range(26)]
s = [False] * 26
cnt = 0
n = len(words)
for i in range(n - 1):
for c in words[i]:
if cnt == 26:
break
o = ord(c) - ord('a')
if not s[o]:
cnt += 1
s[o] = True
m = len(words[i])
for j in range(m):
if j >= len(words[i + 1]):
return ''
c1, c2 = words[i][j], words[i + 1][j]
if c1 == c2:
continue
o1, o2 = ord(c1) - ord('a'), ord(c2) - ord('a')
if g[o2][o1]:
return ''
g[o1][o2] = True
break
for c in words[n - 1]:
if cnt == 26:
break
o = ord(c) - ord('a')
if not s[o]:
cnt += 1
s[o] = True
indegree = [0] * 26
for i in range(26):
for j in range(26):
if i != j and s[i] and s[j] and g[i][j]:
indegree[j] += 1
q = deque()
ans = []
for i in range(26):
if s[i] and indegree[i] == 0:
q.append(i)
while q:
t = q.popleft()
ans.append(chr(t + ord('a')))
for i in range(26):
if s[i] and i != t and g[t][i]:
indegree[i] -= 1
if indegree[i] == 0:
q.append(i)
return '' if len(ans) < cnt else ''.join(ans)class Solution {
public String alienOrder(String[] words) {
boolean[][] g = new boolean[26][26];
boolean[] s = new boolean[26];
int cnt = 0;
int n = words.length;
for (int i = 0; i < n - 1; ++i) {
for (char c : words[i].toCharArray()) {
if (cnt == 26) {
break;
}
c -= 'a';
if (!s[c]) {
++cnt;
s[c] = true;
}
}
int m = words[i].length();
for (int j = 0; j < m; ++j) {
if (j >= words[i + 1].length()) {
return "";
}
char c1 = words[i].charAt(j), c2 = words[i + 1].charAt(j);
if (c1 == c2) {
continue;
}
if (g[c2 - 'a'][c1 - 'a']) {
return "";
}
g[c1 - 'a'][c2 - 'a'] = true;
break;
}
}
for (char c : words[n - 1].toCharArray()) {
if (cnt == 26) {
break;
}
c -= 'a';
if (!s[c]) {
++cnt;
s[c] = true;
}
}
int[] indegree = new int[26];
for (int i = 0; i < 26; ++i) {
for (int j = 0; j < 26; ++j) {
if (i != j && s[i] && s[j] && g[i][j]) {
++indegree[j];
}
}
}
Deque<Integer> q = new LinkedList<>();
for (int i = 0; i < 26; ++i) {
if (s[i] && indegree[i] == 0) {
q.offerLast(i);
}
}
StringBuilder ans = new StringBuilder();
while (!q.isEmpty()) {
int t = q.pollFirst();
ans.append((char) (t + 'a'));
for (int i = 0; i < 26; ++i) {
if (i != t && s[i] && g[t][i]) {
if (--indegree[i] == 0) {
q.offerLast(i);
}
}
}
}
return ans.length() < cnt ? "" : ans.toString();
}
}class Solution {
public:
string alienOrder(vector<string>& words) {
vector<vector<bool>> g(26, vector<bool>(26));
vector<bool> s(26);
int cnt = 0;
int n = words.size();
for (int i = 0; i < n - 1; ++i) {
for (char c : words[i]) {
if (cnt == 26) break;
c -= 'a';
if (!s[c]) {
++cnt;
s[c] = true;
}
}
int m = words[i].size();
for (int j = 0; j < m; ++j) {
if (j >= words[i + 1].size()) return "";
char c1 = words[i][j], c2 = words[i + 1][j];
if (c1 == c2) continue;
if (g[c2 - 'a'][c1 - 'a']) return "";
g[c1 - 'a'][c2 - 'a'] = true;
break;
}
}
for (char c : words[n - 1]) {
if (cnt == 26) break;
c -= 'a';
if (!s[c]) {
++cnt;
s[c] = true;
}
}
vector<int> indegree(26);
for (int i = 0; i < 26; ++i)
for (int j = 0; j < 26; ++j)
if (i != j && s[i] && s[j] && g[i][j])
++indegree[j];
queue<int> q;
for (int i = 0; i < 26; ++i)
if (s[i] && indegree[i] == 0)
q.push(i);
string ans = "";
while (!q.empty()) {
int t = q.front();
ans += (t + 'a');
q.pop();
for (int i = 0; i < 26; ++i)
if (i != t && s[i] && g[t][i])
if (--indegree[i] == 0)
q.push(i);
}
return ans.size() < cnt ? "" : ans;
}
};