Circular Local MiniMax
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given n integers a1,a2,…,an. Is it possible to arrange them on a circle so that each number is strictly greater than both its neighbors or strictly smaller than both its neighbors?

In other words, check 
 
if there exists a rearrangement b1,b2,…,bn of the integers a1,a2,…,an such that for each i from 1 to n at least one of the following conditions holds:

bi−1<bi>bi+1
bi−1>bi<bi+1
To make sense of the previous formulas for i=1 and i=n, one shall define b0=bn and bn+1=b1.

Input
The first line of the input contains a single integer t (
 
1≤t≤3⋅104)  — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer n (3≤n≤105)  — the number of integers.

The second line of each test case contains n integers a1,a2,…,an (0≤ai≤109).

The sum of n over all test cases doesn't exceed 2⋅105.
 


Output
For each test case, if it is not possible to arrange the numbers on the circle satisfying the conditions from the statement, output NO. You can output each letter in any case.

Otherwise, output YES. In the second line, output n integers b1,b2,…,bn, which are a rearrangement of a1,a2,…,an and satisfy the conditions from the statement. If there are multiple valid ways to arrange the numbers, you can output any of them.

Example
inputCopy
4
3
1 1 2
4
1 9 8 4
4
2 0 2 2
6
1 1 1 11 111 1111
outputCopy
NO
YES
1 8 4 9 
NO
YES
1 11 1 111 1 1111 
Note
It can be shown that there are no valid arrangements for the first and the third test cases.

In the second test case, the arrangement [1,8,4,9] works. In this arrangement, 1 and 4 are both smaller than their neighbors, and 8,9 are larger.

In the fourth test case, the arrangement [1,11,1,111,1,1111] works. In this arrangement, the three elements equal to 1 are smaller than their neighbors, while all other elements are larger than their neighbors.

思路：

• 首先奇數數列，一定不會滿足，（自己舉一個栗子就行了）
• 偶數數列： 直接貪心 排序預處理，然後 後面n/2分別放在裡面就行了

#include <bits/stdc++.h>
using namespace std;
#define ri register int
#define  M 100005
// 15:33
template <class G> void read(G &x)
{
    x=0;int f=0;char ch=getchar();
    while(ch<'0'||ch>'9'){f=ch=='-';ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
    x=f?-x:x;
    return ;
}


int n;
int t;
int val[M];
int main(){
    
    read(t);
    while(t--)
    {
        read(n);
        for(ri i=1;i<=n;i++)
        {
            read(val[i]);
        }
        if(n&1)
        {
            printf("NO\n");
            continue;
        }
        sort(val+1,val+1+n);
        int flag=1;
        for(ri i=1;i<=n/2;i++)
        {
            int j=i+(n/2);
            int k=i+1;if(k>(n/2)) k=1;
            if(val[j]<=val[i]||val[j]<=val[k])
            {
                flag=0;break;
            }
        }
        if(flag)
        {
            printf("YES\n");
            for(ri i=1;i<=n/2;i++)
            {
              printf("%d %d ",val[i],val[i+n/2]);
            }
            printf("\n");
        }
        else printf("NO\n");
    }
    
}