Codeforces Round #591 D. Sequence Sorting（逆向思维）

题目链接

https://codeforces.com/contest/1241/problem/D

题目描述

You are given a sequence a1,a2,…,an, consisting of integers.

You can apply the following operation to this sequence: choose some integer x and move all elements equal to x either to the beginning, or to the end of a. Note that you have to move all these elements in one direction in one operation.

For example, if a=[2,1,3,1,1,3,2], you can get the following sequences in one operation (for convenience, denote elements equal to x as x-elements):

• [1,1,1,2,3,3,2] if you move all 1-elements to the beginning;
• [2,3,3,2,1,1,1] if you move all 1-elements to the end;
• [2,2,1,3,1,1,3] if you move all 2-elements to the beginning;
• [1,3,1,1,3,2,2] if you move all 2-elements to the end;
• [3,3,2,1,1,1,2] if you move all 3-elements to the beginning;
• [2,1,1,1,2,3,3] if you move all 3-elements to the end;

You have to determine the minimum number of such operations so that the sequence a becomes sorted in non-descending order. Non-descending order means that for all i from 2 to n, the condition ai−1≤ai is satisfied.

Note that you have to answer q independent queries.

Input

The first line contains one integer q (1≤q≤3⋅10^5) — the number of the queries. Each query is represented by two consecutive lines.

The first line of each query contains one integer n (1≤n≤3⋅10^5) — the number of elements.

The second line of each query contains n integers a1,a2,…,an (1≤ai≤n) — the elements.

It is guaranteed that the sum of all nn does not exceed 3⋅10^5.

Output

For each query print one integer — the minimum number of operation for sorting sequence aa in non-descending order.

Example

input

3
7
3 1 6 6 3 1 1
8
1 1 4 4 4 7 8 8
7
4 2 5 2 6 2 7

output

2
0
1

Note

In the first query, you can move all 1-elements to the beginning (after that sequence turn into [1,1,1,3,6,6,3]) and then move all 6-elements to the end.

In the second query, the sequence is sorted initially, so the answer is zero.

In the third query, you have to move all 2-elements to the beginning.

题解

这题从正面想难度很大，基本上是无解。

我们可以将所有数分为两类，一类需要移动才能将整个序列排为有序，另一类原地不动即可。

可以看出，对那些需要移动的数，只需移动一次即可使数字就位。

那么只需求出最多有多少种数字可以原地不动就行了。

这样这道题难度就不大了。

代码

#include<bits/stdc++.h>
#define ll long long 
#define maxn 300010
using namespace std;
void read(ll &x){
    ll f=1;x=0;char s=getchar();
    while(s<'0'||s>'9'){if(s=='-')f=-1;s=getchar();}
    while(s>='0'&&s<='9'){x=x*10+s-'0';s=getchar();}
    x*=f;
}
 
ll t,n,f[maxn],ns[maxn];
ll a[maxn],l[maxn],r[maxn];
vector<ll>st; 
int main(){
	read(t);
	ll i;
	while(t--){
		read(n);
		for(i=1;i<=n;i++)read(a[i]);
		st.clear();
		for(i=0;i<=n+1;i++){
			f[i]=0;
			l[i]=0;
			r[i]=0;
			ns[i]=0;
		}
		for(i=1;i<=n;i++){
			if(!l[a[i]]){
				l[a[i]]=i;
				r[a[i]]=i;
				//st.push_back(a[i]);
			}
			else{
				l[a[i]]=min(l[a[i]],i);
				r[a[i]]=max(r[a[i]],i);
			}
		}
		ll be=0;
		for(i=1;i<=n;i++){
			if(l[i]){
				st.push_back(i);
				ns[i]=be;
				be=i;
			}
		}
		f[st[0]]=1;
		for(i=1;i<st.size();i++){
			int v=st[i],u=st[i-1];
			f[v]=1;
			if(u==ns[v]&&l[v]>r[u])f[v]=f[u]+1;
		}
		ll ans=0;
		for(i=1;i<=n;i++){
			ans=max(f[i],ans);
		}
		cout<<st.size()-ans<<endl;
	}
	
	
	return 0;
}