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 namespace std;
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t = 1;
    cin >> t;
    while (t--) {
      string s;
      cin >> s;
      int k;
      cin >> k;
      vector<int> a(10, 0);
      int k1 = k;
      int pos = s.size();
      for (int i = 0; i < s.size(); ++i) {
        a[s[i] - '0']++;
        if (a[s[i] - '0'] == 1) {
          --k1;
        }
        if (k1 == -1) {
          a[s[i] - '0']--;
          pos = i;
          break;
        }
      }
      if (pos == s.size()) {
        cout << s << '\n';
        continue;
      }
      int ch = -1;
      char smth;
      int zero = 0;
      bool ok = 0;
      for (int i = s[pos] - '0' + 1; i <= 9; ++i) {
        if (a[i] > 0) {
          ch = pos;
          smth = i + '0';
          ok = 1;
          break;
        }
      }
      if (!ok) {
        for (int i = pos - 1; i >= 0; --i) {
          if (a[s[i] - '0'] == 1 && s[i] != '9') {
            ch = i;
            a[s[i] - '0' + 1]++;
            smth = s[i] + 1;
            ok = 1;
          } else
            for (int j = s[i] - '0' + 1; j <= 9; ++j) {
              if (a[j] > 0) {
                ch = i;
                smth = j + '0';
                ok = 1;
                break;
              }
            }
          a[s[i] - '0']--;
          if (ok) break;
        }
      }
      if (ch != -1) {
        char minn;
        s[ch] = smth;
        int cnt = 0;
        for (int i = 0; i < 10; ++i) {
          if (a[i] > 0) ++cnt;
        }
        for (int i = 0; i < 10; ++i) {
          if (a[i] > 0) {
            minn = i + '0';
            break;
          }
        }
        if (cnt < k) minn = '0';
        for (int i = ch + 1; i < s.size(); ++i) {
          s[i] = minn;
        }
        cout << s << '\n';
      }
    }
  }
  