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: Wrong Answer)

/*
	Author: Ritik Patel
*/
 
 
//&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& STL DEBUGGER &&&&&&&&&&&&&&&&&&&&&&&&&&&
 
// #define _GLIBCXX_DEBUG       // Iterator safety; out-of-bounds access for Containers, etc.
// #pragma GCC optimize "trapv" // abort() on (signed) integer overflow.
 
//&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& LIBRARIES &&&&&&&&&&&&&&&&&&&&&&&&&&&
 
#include <bits/stdc++.h>
using namespace std;
 
/*#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp> // Including tree_order_statistics_node_update
template<typename T, typename V = __gnu_pbds::null_type>
using ordered_set = __gnu_pbds::tree<T, V, less<T>, __gnu_pbds::rb_tree_tag, __gnu_pbds::tree_order_statistics_node_update>; 
*/
//find_by_order()->returns an iterator to the k-th largest element(0-based indexing)
//order_of_key()->Number of items that are strictly smaller than our item
 
//&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& DEFINES &&&&&&&&&&&&&&&&&&&&&&&&&&&
 
#define int long long int
// #define ll long long int
#define all(i) i.begin(), i.end()
#define sz(a) (int)a.size()
 
// #define ld long double
// const ld PI  = 3.141592;
const int dx4[4] = {0, 1, 0, -1};
const int dy4[4] = {-1, 0, 1, 0};
const int dx8[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
const int dy8[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
 
//&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& DEBUG &&&&&&&&&&&&&&&&&&&&&&&&&&&
 
#define XOX
vector<string> vec_splitter(string s) {
    for(char& c: s) c = c == ','?  ' ': c;
    stringstream ss; ss << s;
    vector<string> res;
    for(string z; ss >> z; res.push_back(z));
    return res;
}
 
void debug_out(vector<string> __attribute__ ((unused)) args, __attribute__ ((unused)) int idx) { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, Head H, Tail... T) {
    if(idx > 0) cerr << ", ";
    stringstream ss; ss << H;
    cerr << args[idx] << " = " << ss.str();
    debug_out(args, idx + 1, T...);
}
 
#ifdef XOX
#define debug(...) debug_out(vec_splitter(#__VA_ARGS__), 0, __VA_ARGS__)
#else
#define debug(...) 42
#endif
 
 
//&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& CODE &&&&&&&&&&&&&&&&&&&&&&&&&&&

int N, K;
const int MAXN = 3e5 + 5;
vector<int> A(MAXN);
int memo[MAXN][2];

int dp(int id, bool taken, int prev) {
    if(id == N) {
        return 0;
    }
    auto &res = memo[id][taken];
    if(res != -1) {
        return res;
    }
    res = -1e9;
    // take this element
    if(prev == -1) {
        res = max(res, dp(id + 1, true, A[id]) + 1);
    } else {
        // take only if we match contraint
        if(abs(prev - A[id]) <= K ){
            res = max(res, dp(id + 1, true, A[id]) + 1);
        }
    }

    // skip this element
    res = max(res, dp(id + 1, false, prev));

    return res;
}

void solve(){
    cin >> N >> K;
    for(int i = 0; i < MAXN; ++i)
        for(int j = 0; j < 2; ++j)
            memo[i][j] = -1;
    for(int i = 0; i < N; ++i) {
        cin >> A[i];
    }
    cout << dp(0, 0, -1);
    
}

int32_t main(){
    // freopen("input.txt","r",stdin);
    // freopen("output.txt","w",stdout); 
    ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
    int T = 1; 
    // cin >> T;
    for(int i = 1; i <= T; ++i){
        // cout << "Case #" << i << ": ";
        solve();
    }
    return 0;
}
 
/*
Sample inp
*/