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 mxN = 2e5 + 5;
  vector<int> a[mxN];
  int tree[4 * mxN];
  void update(int x, int lx, int rx, int i, int v) {
    if (lx == rx) {
      tree[x] = max(tree[x], v);
      return;
    }
    int mx = (lx + rx) >> 1;
    if (i <= mx)
      update(x << 1, lx, mx, i, v);
    else
      update(x << 1 | 1, mx + 1, rx, i, v);
    tree[x] = max(tree[x << 1], tree[x << 1 | 1]);
  }
  int query(int x, int lx, int rx, int l, int r) {
    if (l > rx || lx > r) return 0;
    if (l <= lx && r >= rx) return tree[x];
    int mx = (lx + rx) >> 1;
    return max(query(x << 1, lx, mx, l, r), query(x << 1 | 1, mx + 1, rx, l, r));
  }
  void testCase() {
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
      int x;
      cin >> x;
      a[x].push_back(i);
    }
    for (int i = 1; i <= n; i++) {
      reverse(a[i].begin(), a[i].end());
      vector<int> ans(a[i].size());
      for (int j = 0; j < a[i].size(); j++) {
        if (i - a[i][j] > 0) continue;
        ans[j] = query(1, 1, n, 1, a[i][j] - i + 1) + 1;
      }
      for (int j = 0; j < a[i].size(); j++) {
        if (i - a[i][j] > 0) continue;
        update(1, 1, n, a[i][j] - i + 1, ans[j]);
      }
    }
    cout << query(1, 1, n, 1, n) << "\n";
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    testCase();
    return 0;
  }
  