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 1.8.0) (judge verdict: Wrong Answer)

import java.math.*;
import java.util.*;
import java.util.function.*;
import java.util.stream.*;

public class Main {
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int k = sc.nextInt();
    List<Integer> list = new ArrayList<>();
    for (int i = 0; i < n; i++) {
      list.add(sc.nextInt());
    }
    map.put(0, 1);
    lastSet.add(0);
    for (int i = 1; i < n; i++) {
      int v = list.get(i);
      int max = 0;
      int maxI = i;
      for (int j : lastSet) {
        int newV = map.get(j);
        boolean use = Math.abs(v - list.get(j)) <= k;
        if (use) {
          newV++;
        }
        if (max < newV) {
          max = newV;
          if (use) {
            maxI = j;
          }
        }
      }
      lastSet.remove(maxI);
      lastSet.add(i);
      map.put(i, max);
    }
    System.out.println(map.values().stream().max(Comparator.naturalOrder()).get());
  }

  static final Map<Integer, Integer> map = new HashMap<>();
  static final Set<Integer> lastSet = new HashSet<>();
//
//  static int culc(List<Integer> list, int i, int limit) {
//    return map.computeIfAbsent(i, k -> {
//      if (i == 0) {
//        lastSet.add(i);
//        return 1;
//      }
//      int v = list.get(i);
//      int max = 0;
//      int maxI = i;
//      for (int j : lastSet) {
//        int newV = culc(list, j, limit);
//        if (Math.abs(v - list.get(j)) <= limit) {
//          newV++;
//        }
//        if (max < newV) {
//          max = newV;
//          maxI = j;
//        }
//      }
//      lastSet.remove(maxI);
//      lastSet.add(i);
//      return max;
//    });
//  }
}