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[] args){
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
        int k = sc.nextInt();
		int [] in = new int [n];  
        for(int i=0;i<n;i++){
                             in[i] = sc.nextInt();
                            }
        int ans = 1;
		for(int j=0;j<n;j++){
        int ans1 = 1;
        int point = j;
        while(point<n){for(int i=point+1;i<n;i++)
        {if(Math.abs(in[i]-in[point])<=k){ans1++;point=i;break;}                                                
         else{if(i==n-1){point=n;break;}                                                     
             }                                                                                                   
        } if(point==n-1){point=n;}
                      } 
        ans = Math.max(ans,ans1);
        }
		System.out.println(ans);
	}
}