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 PyPy3 (7.3.0) (judge verdict: Accepted)

def resolve():
    N, K = map(int, input().split())
    n= 300010
    #####segfunc######
    def segfunc(x, y):
        return max(x, y)

    def init(init_val):
        # set_val
        for i in range(n):
            seg[i + num - 1] = init_val[i]
            # built
        for i in range(num - 2, -1, -1):
            seg[i] = segfunc(seg[2 * i + 1], seg[2 * i + 2])

    def update(k, x):
        k += num - 1
        seg[k] = x
        while k:
            k = (k - 1) // 2
            seg[k] = segfunc(seg[k * 2 + 1], seg[k * 2 + 2])

    def query(p, q):
        if q <= p:
            return ide_ele
        p += num - 1
        q += num - 2
        res = ide_ele
        while q - p > 1:
            if p & 1 == 0:
                res = segfunc(res, seg[p])
            if q & 1 == 1:
                res = segfunc(res, seg[q])
                q -= 1
            p = p // 2
            q = (q - 1) // 2
        if p == q:
            res = segfunc(res, seg[p])
        else:
            res = segfunc(segfunc(res, seg[p]), seg[q])
        return res

    #####単位元######
    ide_ele = 0

    # num:n以上の最小の2のべき乗
    num = 2 ** (n - 1).bit_length()
    seg = [ide_ele] * 2 * num

    #init([0]*300005)

    for i in range(N):
        a = int(input())
        b = query(max(0, a-K), min(a+K+1, 300001))
        update(a, b+1)
    print(query(0, 300000))
resolve()