There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of non-empty words from the dictionary, where words are sorted lexicographically by the rules of this new language. Derive the order of letters in this language.

Note:

  1. You may assume all letters are in lowercase.
  2. You may assume that if a is a prefix of b, then a must appear before b in the given dictionary.
  3. If the order is invalid, return an empty string.
  4. There may be multiple valid order of letters, return any one of them is fine.

Test Cases

Examples 1:

Input:
[
  "wrt",
  "wrf",
  "er",
  "ett",
  "rftt"
]

Output: "wertf"

Examples 2:

Input:
[
  "z",
  "x"
]
Output: "zx"

Examples 3:

Input:
[
  "z",
  "x",
  "z"
]
Output: ""
Explanation: The order is invalid, so return "".

Constraints:

Solution

class Solution {
public:
    string alienOrder(vector<string>& A) {
        unordered_map<int, unordered_set<int>> G;
        int indegree[26] = {};
        for (auto &s : A) {
            for (char c : s) G[c - 'a'] = {};
        }
        for (int i = 1; i < A.size(); ++i) {
            int j = 0;
            for (; j < min(A[i - 1].size(), A[i].size()); ++j) {
                if (A[i - 1][j] == A[i][j]) continue;
                G[A[i - 1][j] - 'a'].insert(A[i][j] - 'a');
                break;
            }
            if (j == A[i].size() && j < A[i - 1].size()) return "";
        }
        for (auto &[from, tos] : G) {
            for (int to : tos) {
                indegree[to]++;
            }
        }
        queue<int> q;
        for (int i = 0; i < 26; ++i) {
            if (G.count(i) && indegree[i] == 0) q.push(i);
        }
        string ans;
        while (q.size()) {
            int u = q.front();
            q.pop();
            ans += u + 'a';
            for (int v : G[u]) {
                if (--indegree[v] == 0) q.push(v);
            }
        }
        return ans.size() == G.size() ? ans : "";
    }
};
class Solution {
    // Returns the order of characters in the alien language, or "" if invalid
    public String alienOrder(String[] words) {
        // Adjacency matrix for graph representation
        boolean[][] g = new boolean[26][26];
        // Set of present characters
        boolean[] s = new boolean[26];
        int cnt = 0;
        int n = words.length;
        // Mark all unique characters and build graph edges
        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 ""; // prefix case
                char c1 = words[i].charAt(j), c2 = words[i + 1].charAt(j);
                if (c1 == c2) continue;
                if (g[c2 - 'a'][c1 - 'a']) return ""; // cycle check
                g[c1 - 'a'][c2 - 'a'] = true;
                break;
            }
        }
        // Mark unique characters in the last word
        for (char c : words[n - 1].toCharArray()) {
            if (cnt == 26) break;
            c -= 'a';
            if (!s[c]) {
                ++cnt;
                s[c] = true;
            }
        }
        // Calculate indegrees
        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];
                }
            }
        }
        // Topological sort using Kahn's algorithm
        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);
                    }
                }
            }
        }
        // If not all characters are used, return ""
        return ans.length() < cnt ? "" : ans.toString();
    }
}
import collections


class Node(object):
  def __init__(self, val):
    self.val = val
    self.neighbors = []

  def connect(self, node):
    self.neighbors.append(node)

  def getNbrs(self):
    return self.neighbors


class Solution(object):
  def alienOrder(self, words):
    """
    :type words: List[str]
    :rtype: str
    """

    def dfs(root, graph, visited):
      visited[root] = 1
      for nbr in graph[root].getNbrs():
        if visited[nbr.val] == 0:
          if not dfs(nbr.val, graph, visited):
            return False
        elif visited[nbr.val] == 1:
          return False

      visited[root] = 2
      self.ans += root
      return True

    self.ans = ""
    graph = {}
    visited = collections.defaultdict(int)
    self.topNum = 0
    for i in range(0, len(words) - 1):
      a = words[i]
      b = words[i + 1]
      i = 0
      while i < len(a) and i < len(b):
        if a[i] != b[i]:
          nodeA = nodeB = None
          if a[i] not in graph:
            nodeA = Node(a[i])
            graph[a[i]] = nodeA
          else:
            nodeA = graph[a[i]]
          if b[i] not in graph:
            nodeB = Node(b[i])
            graph[b[i]] = nodeB
          else:
            nodeB = graph[b[i]]
          nodeA.connect(nodeB)
          break
        i += 1
      if i < len(a) and i >= len(b):
        return ""

    for c in graph:
      if visited[c] == 0:
        if not dfs(c, graph, visited):
          return ""

    unUsedSet = set()
    for word in words:
      for c in word:
        unUsedSet.add(c)

    for c in unUsedSet:
      if c not in graph:
        self.ans += c
    return self.ans[::-1]
Time Complexity: O(n)
Space Complexity: O(1)