description: It is a complicated version of problem F1. The difference between them is the constraints (F1: k ≤ 2, F2: k ≤ 10).
  
  You are given an integer n. Find the minimum integer x such that x ≥ n and the number x is k-beautiful.
  
  A number is called k-beautiful if its decimal representation having no leading zeroes contains no more than k different digits. E.g. if k = 2, the numbers 3434443, 55550, 777 and 21 are k-beautiful whereas the numbers 120, 445435 and 998244353 are not.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
  
  Each test case consists of one line containing two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 10).
  
  Output
  
  For each test case output on a separate line x — the minimum k-beautiful integer such that x ≥ n.
  
  Example
  
  Input
  
  
  6
  2021 3
  177890 2
  34512 3
  724533 4
  998244353 1
  12345678 10
  
  
  Output
  
  
  2021
  181111
  34533
  724542
  999999999
  12345678
solution: #include <bits/stdc++.h>
  using ll = long long;
  using ld = long double;
  using ull = unsigned long long;
  using namespace std;
  int T, K;
  string n;
  bool use[10];
  bool smaller(const string &s, const string &t) {
    if (t.empty()) return 1;
    for (int i = (0); i <= (((int)(s).size()) - 1); ++i)
      if (s[i] < t[i])
        return 1;
      else if (t[i] < s[i])
        return 0;
    return 0;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> T;
    while (T--) {
      cin >> n >> K;
      for (int i = (0); i <= (9); ++i) use[i] = 0;
      int num = 0;
      string ans = "", cur = "";
      for (int i = (0); i <= (((int)(n).size()) - 1); ++i) {
        int mn = n[i] - '0' + 1;
        if (num == K) {
          mn = 10;
          for (int j = (n[i] - '0' + 1); j <= (9); ++j)
            if (use[j]) {
              mn = j;
              break;
            }
        }
        if (mn <= 9) {
          int add = !use[mn];
          if (num + add < K) {
            string tem = cur;
            tem += (mn + '0');
            while (((int)(tem).size()) < ((int)(n).size())) tem += '0';
            if (smaller(tem, ans)) ans = tem;
          } else {
            int mn1 = mn;
            for (int j = (0); j <= (mn - 1); ++j)
              if (use[j]) {
                mn1 = j;
                break;
              }
            string tem = cur;
            tem += (mn + '0');
            while (((int)(tem).size()) < ((int)(n).size()))
              tem += (char)(mn1 + '0');
            if (smaller(tem, ans)) ans = tem;
          }
        }
        if (!use[n[i] - '0']) num++, use[n[i] - '0'] = 1;
        if (num > K) break;
        cur += n[i];
      }
      if (num <= K) ans = cur;
      cout << ans << '\n';
    }
    return 0;
  }
  