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 Java (OpenJDK 11.0.6) (judge verdict: Time Limit Exceeded)

import java.util.*;
public class Main{
 public static void main(String arg[])
 {
   Scanner sc=new Scanner(System.in);
    
   int n=sc.nextInt();
   int k=sc.nextInt();
   
   int a[]=new int[n];
   
   for(int i=0;i<n;i++)
   {
       a[i]=sc.nextInt();
       
   }
   int s[]=new int[n];
   
   Arrays.fill(s, 1);
   
   for(int i=1;i<n;i++)
   {
       int j=i-1;
       
       while(j>=0)
       {
           if(Math.abs(a[i]-a[j])<=k)
           {
               s[i]=Math.max(s[i],s[j]+1);
           }
           j--;
       }
   }
   int max=0;
   for(int i=0;i<n;i++)max=Math.max(max,s[i]);
     System.out.println(max);
   
 }
}