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>
  #pragma GCC optimize("O3")
  using namespace std;
  int main() {
    int n;
    cin >> n;
    vector<int> v(n);
    vector<pair<int, int>> vec;
    for (int i = 0; i < n; i++) {
      cin >> v[i];
      int orig = v[i];
      v[i] = (i + 1) - v[i];
      if (v[i] >= 0) {
        vec.push_back(make_pair(v[i], orig));
      }
    }
    sort(vec.begin(), vec.end());
    vector<int> pos;
    for (pair<int, int> i : vec) {
      pos.push_back(i.second);
    }
    vector<int> ans;
    n = pos.size();
    for (int i = 0; i < n; i++) {
      auto it = lower_bound(ans.begin(), ans.end(), pos[i]);
      if (it == ans.end()) {
        ans.push_back(pos[i]);
      } else {
        *it = pos[i];
      }
    }
    cout << ans.size() << endl;
  }
  