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 const N = 2e5 + 123;
  vector<int> ST(4 * N, 0);
  void update(int k, int l, int r, int idx, int val) {
    if (l > idx || r < idx) return;
    if (l == r && l == idx) {
      ST[k] = val;
      return;
    }
    int m = (l + r) >> 1;
    update(k << 1, l, m, idx, val);
    update(k << 1 | 1, m + 1, r, idx, val);
    ST[k] = max(ST[k << 1], ST[k << 1 | 1]);
  }
  int getLis(int k, int l, int r, int from, int to) {
    if (l > r) return 0;
    if (l > to || from > r) return 0;
    if (from <= l && r <= to) return ST[k];
    int m = (l + r) >> 1;
    return max(getLis(k << 1, l, m, from, to),
               getLis(k << 1 | 1, m + 1, r, from, to));
  }
  void TestCase() {
    int n;
    cin >> n;
    vector<int> v(n);
    for (auto& x : v) cin >> x;
    vector<pair<int, int>> p;
    for (int i = 0; i < n; i++) {
      if (i - v[i] + 1 >= 0) {
        p.push_back({i - v[i] + 1, v[i]});
      }
    }
    sort(p.begin(), p.end());
    vector<int> LIS(2 * N + 5, 0);
    int answer = 0;
    for (int i = 0; i < p.size(); i++) {
      LIS[p[i].second] = getLis(1, 0, n, 0, p[i].second - 1) + 1;
      update(1, 0, n, p[i].second, LIS[p[i].second]);
      answer = max(answer, LIS[p[i].second]);
    }
    cout << answer << '\n';
  }
  int main() {
    ios_base ::sync_with_stdio(false);
    cin.tie();
    cout.tie();
    int T = 1;
    while (T--) {
      TestCase();
    }
  }
  