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const int N = 26, M = N * N;

class Solution {
 public:
  int cnt = 0;
  bool st[N];
  int h[N], e[M], ne[M], in[N], out[N], idx;
  void add(int a, int b) {
    e[idx] = b;
    ne[idx] = h[a];
    h[a] = idx++;
    out[a]++;
    in[b]++;
  }
  string alienOrder(vector<string>& words) {
    memset(h, -1, sizeof h);
    memset(st, 0, sizeof st);
    int n = words.size();
    for (int i = 0; i < n; i++) {
      for (auto& c : words[i]) {
        if (!st[c - 'a'] && ++cnt >= 0) st[c - 'a'] = true;
      }
      for (int j = 0; j < i; j++) {
        if (!build(words[j], words[i])) return "";
      }
    }
    queue<int> q;
    // 将入度为 0 的节点放入队列
    for (int i = 0; i < 26; i++) {
      if (st[i] && in[i] == 0) q.push(i);
    }
    string s;
    // 拓扑排序
    while (q.size()) {
      int u = q.front();
      q.pop();
      s += u + 'a';
      for (int i = h[u]; i != -1; i = ne[i]) {
        int j = e[i];
        if (--in[j] == 0) q.push(j);
      }
    }
    return s.size() == cnt ? s : "";
  }
  bool build(string a, string b) {
    int n = a.size(), m = b.size(), len = min(n, m);
    for (int i = 0; i < len; i++) {
      int c1 = a[i] - 'a', c2 = b[i] - 'a';
      if (c1 != c2) {
        add(c1, c2);
        return true;
      }
    }
    return n <= m;
  }
};