The only difference between easy and hard versions are constraints on n and k.
You are messaging in one of the popular social networks via your smartphone. Your smartphone can show at most k most recent conversations with your friends. Initially, the screen is empty (i.e. the number of displayed conversations equals 0).
Each conversation is between you and some of your friends. There is at most one conversation with any of your friends. So each conversation is uniquely defined by your friend.
You (suddenly!) have the ability to see the future. You know that during the day you will receive n messages, the i-th message will be received from the friend with ID idi (1≤idi≤109).
If you receive a message from idi in the conversation which is currently displayed on the smartphone then nothing happens: the conversations of the screen do not change and do not change their order, you read the message and continue waiting for new messages.
Otherwise (i.e. if there is no conversation with idi on the screen):
Firstly, if the number of conversations displayed on the screen is k, the last conversation (which has the position k) is removed from the screen.
Now the number of conversations on the screen is guaranteed to be less than k and the conversation with the friend idi is not displayed on the screen.
The conversation with the friend idi appears on the first (the topmost) position on the screen and all the other displayed conversations are shifted one position down.
Your task is to find the list of conversations (in the order they are displayed on the screen) after processing all n messages.
Input
The first line of the input contains two integers n and k (1≤n,k≤2⋅105) — the number of messages and the number of conversations your smartphone can show.
The second line of the input contains n integers id1,id2,…,idn (1≤idi≤109), where idi is the ID of the friend which sends you the i-th message.
Output
In the first line of the output print one integer m (1≤m≤min(n,k)) — the number of conversations shown after receiving all n messages.
In the second line print m integers ids1,ids2,…,idsm, where idsi should be equal to the ID of the friend corresponding to the conversation displayed on the position i after receiving all n messages.
Examples
Input
7 2
1 2 3 2 1 3 2
Output
2
2 1
Input
10 4
2 3 3 1 1 2 1 2 3 3
Output
3
1 3 2
Note
In the first example the list of conversations will change in the following way (in order from the first to last message):
[];
[1];
[2,1];
[3,2];
[3,2];
[1,3];
[1,3];
[2,1].
In the second example the list of conversations will change in the following way:
[];
[2];
[3,2];
[3,2];
[1,3,2];
and then the list will not change till the end.

手写队列配合map方便判断某个id是否在队内
map数组存某个id在手写队列的位置
然后依据题意模拟：
新来的在队内什么都不做
新来的不在队内：加入队，map改一下位置，队列长度过长就移动。
写了快一个小时都没写出来我***的垃圾

#include<bits/stdc++.h>
using namespace std;
int q[1000100],head,tot;
map<int,int> mp;//id为x的在哪 
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int n,k,x,cnt;
    cin>>n>>k;
    head=1,tot=0,cnt=0;
    for(int o=1;o<=n;++o)
    {
        cin>>x;
        //cout<<x<<" "<<endl;
        if(mp[x]<head || mp[x]>tot)//x buzai 
        {
            q[++tot]=x;
            mp[x]=tot;
            ++cnt;
            //cout<<x<<" "<<endl;
            if(cnt>k)
            {
                while(q[head]<1) ++head;
                ++head;
                --cnt;
            }
        }
        else//x zai
        {
            /*q[++tot]=x;
            q[mp[x]]=-1;
            mp[x]=tot;*/
        }
        //for(int i=1;i<=n;++i) cout<<q[i]<<' ';cout<<head<<' '<<tot<<' '<<x<<' '<<mp[x]<<endl;
    }

    cout<<cnt<<endl;
    for(int i=tot;i>=head;--i)
    {
        if(q[i]>0) printf("%d ",q[i]);
    }
    return 0;
}