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;
  pair<int, int> pr[233333];
  bool cmp(const pair<int, int> &a, const pair<int, int> &b) {
    if (a.first == b.first)
      return a.second > b.second;
    else
      return a.first < b.first;
  }
  int f[233333], st[233333], top;
  int main() {
    ios_base::sync_with_stdio(false);
    int n, x;
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> x;
      pr[i] = make_pair(x, i - x);
    }
    sort(pr + 1, pr + n + 1, cmp);
    for (int i = 1; i <= n; i++) {
      if (pr[i].second < 0) continue;
      int l = 0, r = top;
      while (l < r) {
        int mid = (l + r + 1) / 2;
        if (pr[i].second >= pr[st[mid]].second)
          l = mid;
        else
          r = mid - 1;
      }
      f[i] = l + 1;
      if (f[i] > top or pr[i].second < pr[st[l + 1]].second) st[l + 1] = i;
      if (f[i] > top) top = f[i];
    }
    cout << top << endl;
    return 0;
  }
  