Problem:
Score : 400 points

Problem Statement

You are given a sequence A_1, A_2, ..., A_N and an integer K.

Print the maximum possible length of a sequence B that satisfies the following conditions:

B is a (not necessarily continuous) subsequence of A.

For each pair of adjacents elements of B, the absolute difference of the elements is at most K.

Constraints

1 \leq N \leq 300,000

0 \leq A_i \leq 300,000

0 \leq K \leq 300,000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
A_1
A_2
:
A_N

Output

Print the answer.

Sample Input 1

10 3
1
5
4
3
8
6
9
7
2
4

Sample Output 1

7

For example, B = (1, 4, 3, 6, 9, 7, 4) satisfies the conditions.

It is a subsequence of A = (1, 5, 4, 3, 8, 6, 9, 7, 2, 4).

All of the absolute differences between two adjacent elements (|1-4|, |4-3|, |3-6|, |6-9|, |9-7|, |7-4|) are at most K = 3.

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include "bits/stdc++.h"
#define lson t<<1,l,mid
#define rson t<<1|1,mid+1,r
#define L t<<1
#define R t<<1|1
using namespace std;
const int N = 3e5+5;
const int h = 1e5+1;

int mx[N<<2];
void update(int t,int l,int r,int pos,int d){
    if(l==r){ mx[t] = d; return; }
    int mid = (l+r)>>1;
    if(pos<=mid) update(lson,pos,d);
    else update(rson,pos,d);
    mx[t] = max( mx[L] , mx[R] );
}

int query(int t,int l,int r,int ei,int ej){
    if(ei<=l && r<=ej) return mx[t];
    int mid = (l+r)>>1;
    if(ej<=mid) return query(lson,ei,ej);
    else if(ei>mid) return query(rson,ei,ej);
    else return max( query(lson,ei,ej) , query(rson,ei,ej) );
}

int main()
{
    ios::sync_with_stdio(false);

    int n,k;  cin>>n>>k;
    int ans = 0;
    for(int i=1;i<=n;i++){
        int a;  cin>>a; a++;
        int b = query(1,1,h, max(a-k,1) , min(a+k,h) )+1;
        ans = max( ans , b );
        update(1,1,h,a,b);
    }
    cout<<ans<<endl;
}