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;
  using ll = long long;
  int n, a[200200];
  set<pair<int, int>> s[200200];
  int main() {
    cin.tie(0)->sync_with_stdio(0);
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    vector<pair<int, int>> v;
    for (int i = 1; i <= n; i++)
      if (a[i] <= i) v.push_back({i - a[i], a[i]});
    sort(v.begin(), v.end());
    for (auto [x, y] : v) {
      int l = 1, r = n;
      while (l <= r) {
        int m = (l + r) / 2;
        auto it = s[m].lower_bound({x, y});
        if (it != s[m].begin() && prev(it)->second < y)
          l = m + 1;
        else
          r = m - 1;
      }
      auto it = next(s[l].insert({x, y}).first);
      while (it != s[l].end() && it->second >= y) it = s[l].erase(it);
    }
    int l = 1;
    while (!s[l].empty()) l++;
    cout << l - 1;
  }
  