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;
  const int NMax = 2e5 + 9;
  pair<int, int> a[NMax];
  int e[NMax];
  bool cmp(pair<int, int> a, pair<int, int> b) {
    if (a.first == b.first) return (a.second > b.second);
    return (a.first < b.first);
  }
  void Solve() {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> a[i].first;
      a[i].second = i - a[i].first + 1;
    }
    sort(a, a + n, cmp);
    int m = 0;
    for (int i = 0; i < n; i++)
      if (a[i].second >= 0) {
        int f = upper_bound(e, e + m, a[i].second) - e;
        if (f == m) m++;
        e[f] = a[i].second;
      }
    cout << m;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    Solve();
  }
  