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;
  long long ans, n, x;
  vector<int> dp;
  vector<pair<int, int>> A;
  int main() {
    cin >> n;
    int curr = 0;
    dp.assign(n + 1, 0);
    A.resize(n + 1);
    A[0] = {-1000000, 0};
    for (int i = 1; i <= n; i++) {
      cin >> x;
      A[i] = {i - x, x};
    }
    sort(A.begin(), A.end());
    for (int i = 1; i <= n; i++) {
      if (A[i].first < 0)
        continue;
      else {
        int val = A[i].second;
        if (val > dp[curr]) {
          curr++;
          dp[curr] = val;
        } else {
          int l = 0, r = curr, mid;
          while (r - l > 1) {
            mid = (l + r) / 2;
            if (val <= dp[mid])
              r = mid;
            else
              l = mid;
          }
          dp[r] = val;
        }
      }
    }
    cout << curr << endl;
  }
  