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 double EPS = 1e-12;
  const int INF = 1e9 + 10;
  template <typename T>
  using MinPQ = priority_queue<T, vector<T>, greater<T>>;
  int n;
  vector<int> a;
  vector<pair<int, int>> pos;
  void Solve() {
    scanf("%d", &n);
    a.resize(n + 1);
    pos.resize(n);
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &a[i]);
      pos[i - 1] = {a[i], i};
    }
    sort((pos).begin(), (pos).end(),
         [&](const pair<int, int>& x, const pair<int, int>& y) -> bool {
           if (x.first == y.first) return x.second > y.second;
           return x.first < y.first;
         });
    int ans = 0;
    map<int, int> f;
    f[0] = 0;
    for (auto [i, p] : pos) {
      int d = p - i;
      if (d < 0) continue;
      int w;
      {
        auto it = prev(f.upper_bound(d));
        w = it->second + 1;
        42;
      }
      ans = max(ans, w);
      f[d] = w;
      auto it = f.upper_bound(d);
      while (it != f.end() && it->second <= w) it = f.erase(it);
    }
    printf("%d\n", ans);
  }
  int main() {
    int t = 1;
    for (int i = 1; i <= t; ++i) {
      Solve();
    }
    return 0;
  }
  