Question

269. Alien Dictionary

中文文档

Description

There is a new alien language that uses the English alphabet. However, the order of the letters is unknown to you.

You are given a list of strings words from the alien language's dictionary. Now it is claimed that the strings in words are sorted lexicographically by the rules of this new language.

If this claim is incorrect, and the given arrangement of string in words cannot correspond to any order of letters, return "".

Otherwise, 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 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 <= 100
• 1 <= words[i].length <= 100
• words[i] consists of only lowercase English letters.

Solutions

Solution 1

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;
  }
};