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;
  int tree[4 * 200005];
  void update(int n, int st, int en, int index, int val) {
    if (st == en) {
      tree[n] = val;
      return;
    }
    int mid = (st + en) / 2, lc = 2 * n, rc = lc + 1;
    if (index <= mid)
      update(lc, st, mid, index, val);
    else
      update(rc, mid + 1, en, index, val);
    tree[n] = max(tree[lc], tree[rc]);
  }
  int query(int n, int st, int en, int l, int r) {
    if (r < st || en < l) return 0;
    if (l <= st && en <= r) return tree[n];
    int mid = (st + en) / 2, lc = 2 * n, rc = lc + 1;
    return max(query(lc, st, mid, l, r), query(rc, mid + 1, en, l, r));
  }
  vector<int> pos[200005 + 5];
  int main() {
    int n, inp;
    scanf("%d", &n);
    int mx = 0;
    for (int i = 1; i < n + 1; i++) {
      scanf("%d", &inp);
      pos[inp].push_back(i);
    }
    for (int num = 1; num < 200005; num++) {
      reverse(pos[num].begin(), pos[num].end());
      for (auto position : pos[num]) {
        if (position < num) continue;
        int highestDeletePossible = position - num;
        int maxAns = query(1, 0, 200005, 0, highestDeletePossible) + 1;
        update(1, 0, 200005, highestDeletePossible, maxAns);
      }
    }
    cout << query(1, 0, 200005, 0, 200005) << '\n';
    return 0;
  }
  