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::sync_with_stdio(0);
    cin.tie(0);
    long long t;
    cin >> t;
    while (t--) {
      long long n, k;
      map<long long, long long> found;
      cin >> n >> k;
      string s = to_string(n);
      long long cnt = 0;
      long long lasttochange = -1;
      long long firstforoccur = -1;
      long long change = -1;
      char min = '9';
      char f = '*';
      for (long long i = 0; i < s.size(); i++) {
        if (!found[s[i]]) {
          cnt++;
          if (cnt == k) firstforoccur = i;
          if (cnt > k) {
            change = i;
            break;
          }
          if (s[i] < min) min = s[i];
        }
        char assign = s[i] + 1;
        while (assign > s[i]) {
          if (found[assign]) {
            lasttochange = i;
            f = assign;
            break;
          }
          if (assign >= '9') {
            break;
          }
          assign++;
        }
        found[s[i]]++;
      }
      if (change == -1)
        cout << s << '\n';
      else {
        char assign = s[change] + 1;
        bool can = false;
        while (assign > s[change]) {
          if (found[assign]) {
            s[change] = assign;
            can = true;
            for (long long i = change + 1; i < s.size(); i++) s[i] = min;
            break;
          }
          if (assign >= '9') {
            break;
          }
          assign++;
        }
        if (!can) {
          if (firstforoccur >= lasttochange) {
            if (s[firstforoccur] == min) min = s[firstforoccur] + 1;
            if (found[s[firstforoccur] + 1]) min = '0';
            s[firstforoccur]++;
            for (long long i = firstforoccur + 1; i < s.size(); i++) {
              s[i] = min;
            }
          } else {
            if (found[s[lasttochange]] == 1) {
              min = '0';
            }
            s[lasttochange] = f;
            for (long long i = lasttochange + 1; i < s.size(); i++) {
              s[i] = min;
            }
          }
        }
        cout << s << '\n';
      }
    }
    return 0;
  }
  