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(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int t, k;
    string n;
    cin >> t;
    vector<int> digits(10);
    while (t-- > 0) {
      cin >> n >> k;
      std::fill(digits.begin(), digits.end(), 0);
      int cnt = 0;
      int up_idx = 1000;
      int i = 0;
      while (i < n.length()) {
        if (i > up_idx) {
          int j = 0;
          while (j < 10) {
            if (digits[j] > 0 || cnt < k) {
              n[i] = (char)(j + '0');
              if (digits[j]++ == 0) cnt++;
              break;
            }
            j++;
          }
        } else {
          int d = n[i] - '0';
          if (digits[d]++ == 0) cnt++;
          if (cnt > k) {
            int j = d + 1;
            digits[d]--;
            while (j < 10) {
              if (digits[j] > 0) {
                n[i] = (char)(j + '0');
                digits[j]++;
                cnt--;
                up_idx = i;
                break;
              }
              j++;
            }
            if (cnt > k) {
              if (--digits[n[i - 1] - '0'] == 0) cnt--;
              cnt--;
              n[i - 1]++;
              up_idx = i - 1;
              i--;
              continue;
            }
          }
        }
        i++;
      }
      cout << n << "\n";
    }
    return 0;
  }
  