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: Time Limit Exceeded)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define ll long long
#define FASTIO ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define inf 1e9
#define ff first
#define ss second
#define lb lower_bound
#define ub upper_bound
#define bs binary_search
#define mod 1000000007
#define pb push_back
#define all(s) s.begin(),s.end()
#define pie 3.14159265358979323846
#define fr(i,a,b) for(ll i=a;i<b;i++)
#define frr(i,a,b) for(ll i=b-1;i>=a;i--)
#define gcd __gcd
#define con continue
#define pii pair<ll,ll>
map<pii,ll>dp;
ll f(ll a[],ll pre,ll n,ll i,ll k){
   if(i==n)return 0;
   ll x=0,y=0;
   if(dp.find({pre,i})!=dp.end())return dp[{pre,a[i]}];
   if(abs(pre-a[i])<=k){
       x=1+f(a,a[i],n,i+1,k);
       y=f(a,pre,n,i+1,k);
   }
   else{
       x=f(a,pre,n,i+1,k);
   }
   dp[{pre,a[i]}]=max(x,y);
   return max(x,y);
}
signed main(){
    FASTIO;
    ll tt=1;
    //cin>>tt;
    while(tt--){
        ll n,k;cin>>n>>k;
        ll a[n];
        for(ll i=0;i<n;i++)cin>>a[i];
        cout<<f(a,a[0],n,0,k);
    }
}