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>
  struct Node {
    int val;
    int idx;
  };
  const int maxn = 2e5 + 10;
  int n;
  Node arr[maxn];
  int b[maxn], cnt;
  int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &arr[i].idx);
      arr[i].val = i - arr[i].idx;
    }
    std::sort(arr + 1, arr + 1 + n, [](const Node &a, const Node &b) {
      if (a.val == b.val) return a.idx < b.idx;
      return a.val < b.val;
    });
    for (int i = 1; i <= n; i++) {
      if (arr[i].val < 0) continue;
      int dex = std::lower_bound(b + 1, b + cnt + 1, arr[i].idx) - b;
      b[dex] = arr[i].idx;
      cnt = std::max(cnt, dex);
    }
    printf("%d\n", cnt);
    return 0;
  }
  