description: Mr. Chanek gives you a sequence a indexed from 1 to n. Define f(a) as the number of indices where a_i = i. 
  
  You can pick an element from the current sequence and remove it, then concatenate the remaining elements together. For example, if you remove the 3-rd element from the sequence [4, 2, 3, 1], the resulting sequence will be [4, 2, 1]. 
  
  You want to remove some elements from a in order to maximize f(a), using zero or more operations. Find the largest possible f(a).
  
  Input
  
  The first line contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the initial length of the sequence.
  
  The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 2 ⋅ 10^5) — the initial sequence a.
  
  Output
  
  Output an integer denoting the largest f(a) that can be obtained by doing zero or more operations.
  
  Examples
  
  Input
  
  
  7
  2 1 4 2 5 3 7
  
  
  Output
  
  
  3
  
  
  Input
  
  
  4
  4 2 3 1
  
  
  Output
  
  
  2
  
  Note
  
  In the first example, f(A) = 3 by doing the following operations.
  
  [2,1,4,2,5,3,7] → [2,1,2,5,3,7] → [1,2,5,3,7] → [1,2,5,3] → [1,2,3]
  
  In the second example, f(A) = 2 and no additional operation is needed.
solution: #include <bits/stdc++.h>
  using namespace std;
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin >> n;
    vector<pair<int, int>> a;
    for (int i = 0; i < n; i++) {
      int val;
      cin >> val;
      int d = i + 1 - val;
      if (d >= 0) {
        a.push_back({d, val});
      }
    }
    sort(a.begin(), a.end());
    vector<int> b(a.size());
    for (int i = 0; i < a.size(); i++) {
      b[i] = a[i].second;
    }
    vector<int> tails;
    for (int i = 0; i < a.size(); i++) {
      int k = tails.size();
      int index =
          upper_bound(tails.begin(), tails.end(), b[i] - 1) - tails.begin() - 1;
      if (k == 0) {
        tails.push_back(b[i]);
      } else if (index == -1) {
        tails[0] = b[i];
      } else if (index == k - 1) {
        if (tails[k - 1] != b[i]) {
          tails.push_back(b[i]);
        }
      } else {
        tails[index + 1] = b[i];
      }
    }
    cout << tails.size() << endl;
  }
  