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 int MAXN = 2e5 + 10;
  int n, a[MAXN];
  pair<int, int> ta[MAXN];
  int seq[MAXN];
  void solve() {
    cin >> n;
    for (int i = (1); i <= (n); ++i) cin >> a[i];
    for (int i = (1); i <= (n); ++i) {
      ta[i].first = i - a[i];
      ta[i].second = a[i];
    }
    sort(ta + 1, ta + 1 + n,
         [](const pair<int, int> &lhs, const pair<int, int> &rhs) {
           return lhs.first < rhs.first ||
                  (lhs.first == rhs.first && lhs.second < rhs.second);
         });
    int i = 1;
    while (i <= n && ta[i].first < 0) ++i;
    if (i > n) {
      cout << "0\n";
      return;
    }
    int ans = 1;
    seq[1] = ta[i].second;
    for (++i; i <= n; ++i) {
      auto p = lower_bound(seq + 1, seq + 1 + ans, ta[i].second);
      *p = ta[i].second;
      ans = max(ans, int(p - seq));
    }
    cout << ans << '\n';
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int _;
    _ = 1;
    while (_--) {
      solve();
    }
    return 0;
  }
  