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>
  template <class T>
  void read(T &a) {
    a = 0;
    char x = getchar();
    bool f = 0;
    for (; x < '0' || x > '9'; x = getchar()) f |= x == '-';
    for (; x >= '0' && x <= '9'; x = getchar()) a = (a << 3) + (a << 1) + x - '0';
    if (f) a = -a;
  }
  template <class T, class... Args>
  void read(T &a, Args &...args) {
    read(a);
    read(args...);
  }
  using namespace std;
  const int N = 2e5 + 5;
  int n;
  int f[N], lim;
  pair<int, int> d[N];
  bool cmp(pair<int, int> a, pair<int, int> b) {
    if (a.first ^ b.first) {
      return a.first < b.first;
    }
    return a.second > b.second;
  }
  int main() {
    read(n);
    for (int i = 1; i <= n; ++i) {
      read(d[i].first);
      d[i].second = i;
    }
    sort(d + 1, d + n + 1, cmp);
    int lim = 0;
    for (int i = 1; i <= n; ++i) {
      int c = d[i].second - d[i].first;
      if (c < 0) continue;
      if (lim == 0) {
        f[++lim] = c;
        continue;
      }
      if (c >= f[lim]) {
        f[++lim] = c;
        continue;
      }
      int l = 1, r = lim + 1;
      while (l < r - 1) {
        int mid = l + r >> 1;
        if (f[mid] <= c) {
          l = mid;
        } else {
          r = mid;
        }
      }
      if (f[l] > c) {
        f[l] = c;
      } else if (l != lim) {
        f[l + 1] = c;
      }
    }
    printf("%d\n", lim);
    return 0;
  }
  