# RPDRDNG - Editorial

Author: S.Manuj Nanthan
Tester : Aryan Choudhary

Easy

# PREREQUISITES:

Maths, Observation

# PROBLEM:

You are given an array B of length 2*N. You have an unknown array A which is sorted and contains distinct elements. You need to find out the array A. B contains all the medians of each prefix and suffix of the array A.

Determine whether B is a valid array of some array A. If yes find the array A else print −1.

# EXPLANATION:

• The median of prefix A[1..1] is equal to A[1]
• The median of prefix A[1..2] is equal to A[1]
• The median of prefix A[1..3] is equal to A[2]

Consider index k,

if k - 1 \le N - k, then the median of A[1...2*k + 1] is equal to A[k] otherwise, the median of A[k - (N - k), N] is equal to A[k]

So for every index k, there exists some prefix or suffix whose median is equal to A[k].

So every element A[k] occurs in array B.

So check if B has N distinct elements. Then we get the array A from these N elements and then construct array B from A and check if it is equal to the given array B.

O(NlogN)

# SOLUTIONS:

Author's Solution
``````#from itertools import *
#from math import *
#from bisect import *
#from collections import *
#from random import *
#from decimal import *
#from heapq import *
#from itertools import *            #Life is hard!
import sys
def inp():
return int(input())
def st():
return input().rstrip('\n')
def lis():
return list(map(int,input().split()))
def ma():
return map(int,input().split())
t=inp()
while(t):
t-=1
n=inp()
a=lis()
if(n==1):
if(a[0]==a[1]):
print(a[0])
else:
print(-1)
elif(n==2):
x=a.count(min(a))
if(x==3):
print(min(a),max(a))
else:
print(-1)
else:
r=sorted(set(a))
maxx=max(r)
xx=len(r)//2
if(len(r)%2==0):
xx-=1
midd=r[xx]
fl=0
dic={}
for i in a:
try:
dic[i]+=1
except:
dic[i]=1
for i in dic.keys():
if(i!=maxx and i!=midd and dic[i]>2):
fl=1
if(dic[maxx]==1 and dic[midd]==3 and fl==0):
print(*r)
else:
print(-1)

``````
Tester's Solution
``````/* in the name of Anton */

/*
Compete against Yourself.
Author - Aryan (@aryanc403)
Atcoder library - https://atcoder.github.io/ac-library/production/document_en/
*/

#ifdef ARYANC403
#else
#pragma GCC optimize ("Ofast")
#pragma GCC target ("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
//#pragma GCC optimize ("-ffloat-store")
#include<bits/stdc++.h>
#define dbg(args...) 42;
#endif

// y_combinator from @neal template https://codeforces.com/contest/1553/submission/123849801
// http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2016/p0200r0.html
template<class Fun> class y_combinator_result {
Fun fun_;
public:
template<class T> explicit y_combinator_result(T &&fun): fun_(std::forward<T>(fun)) {}
template<class ...Args> decltype(auto) operator()(Args &&...args) { return fun_(std::ref(*this), std::forward<Args>(args)...); }
};
template<class Fun> decltype(auto) y_combinator(Fun &&fun) { return y_combinator_result<std::decay_t<Fun>>(std::forward<Fun>(fun)); }

using namespace std;
#define fo(i,n)   for(i=0;i<(n);++i)
#define repA(i,j,n)   for(i=(j);i<=(n);++i)
#define repD(i,j,n)   for(i=(j);i>=(n);--i)
#define all(x) begin(x), end(x)
#define sz(x) ((lli)(x).size())
#define pb push_back
#define mp make_pair
#define X first
#define Y second
#define endl "\n"

typedef long long int lli;
typedef long double mytype;
typedef pair<lli,lli> ii;
typedef vector<ii> vii;
typedef vector<lli> vi;

const auto start_time = std::chrono::high_resolution_clock::now();
void aryanc403()
{
#ifdef ARYANC403
auto end_time = std::chrono::high_resolution_clock::now();
std::chrono::duration<double> diff = end_time-start_time;
cerr<<"Time Taken : "<<diff.count()<<"\n";
#endif
}

long long readInt(long long l, long long r, char endd) {
long long x=0;
int cnt=0;
int fi=-1;
bool is_neg=false;
while(true) {
char g=getchar();
if(g=='-') {
assert(fi==-1);
is_neg=true;
continue;
}
if('0'<=g&&g<='9') {
x*=10;
x+=g-'0';
if(cnt==0) {
fi=g-'0';
}
cnt++;
assert(fi!=0 || cnt==1);
assert(fi!=0 || is_neg==false);

assert(!(cnt>19 || ( cnt==19 && fi>1) ));
} else if(g==endd) {
if(is_neg) {
x=-x;
}
assert(l<=x&&x<=r);
return x;
} else {
assert(false);
}
}
}
string readString(int l, int r, char endd) {
string ret="";
int cnt=0;
while(true) {
char g=getchar();
assert(g!=-1);
if(g==endd) {
break;
}
cnt++;
ret+=g;
}
assert(l<=cnt&&cnt<=r);
return ret;
}
long long readIntSp(long long l, long long r) {
}
long long readIntLn(long long l, long long r) {
}
string readStringLn(int l, int r) {
}
string readStringSp(int l, int r) {
}

assert(getchar()==EOF);
}

vi a(n);
for(int i=0;i<n-1;++i)
return a;
}

// #include<atcoder/dsu>
//     vector<vi> e(n);
//     atcoder::dsu d(n);
//     for(lli i=1;i<n;++i){
//         e[u].pb(v);
//         e[v].pb(u);
//         d.merge(u,v);
//     }
//     assert(d.size(0)==n);
//     return e;
// }

const lli INF = 0xFFFFFFFFFFFFFFFL;

lli seed;
inline lli rnd(lli l=0,lli r=INF)
{return uniform_int_distribution<lli>(l,r)(rng);}

class CMP
{public:
bool operator()(ii a , ii b) //For min priority_queue .
{    return ! ( a.X < b.X || ( a.X==b.X && a.Y <= b.Y ));   }};

void add( map<lli,lli> &m, lli x,lli cnt=1)
{
auto jt=m.find(x);
if(jt==m.end())         m.insert({x,cnt});
else                    jt->Y+=cnt;
}

void del( map<lli,lli> &m, lli x,lli cnt=1)
{
auto jt=m.find(x);
if(jt->Y<=cnt)            m.erase(jt);
else                      jt->Y-=cnt;
}

bool cmp(const ii &a,const ii &b)
{
return a.X<b.X||(a.X==b.X&&a.Y<b.Y);
}

const lli mod = 1000000007L;
// const lli maxN = 1000000007L;

lli T,n,i,j,k,in,cnt,l,r,u,v,x,y;
lli m;
string s;
vi a;
//priority_queue < ii , vector < ii > , CMP > pq;// min priority_queue .

int main(void) {
ios_base::sync_with_stdio(false);cin.tie(NULL);
// freopen("txt.in", "r", stdin);
// freopen("txt.out", "w", stdout);
// cout<<std::fixed<<std::setprecision(35);
const auto solve=[](const vi &a){
const int n=sz(a)/2;
auto b=a;
sort(all(b));
(b).erase(unique(all(b)),(b).end());
if(sz(b)!=n){
cout<<-1<<endl;
return;
}
if(n==1){
cout<<b[0]<<endl;
return;
}
map<lli,lli> mm;
for(auto x:a)
mm[x]++;
for(int i=0;i<n;i++){
const lli x=b[i];
if(i==(n-1)/2){
if(mm[x]!=3){
cout<<-1<<endl;
return;
}
continue;
}
if(i==n-1){
if(mm[x]!=1){
cout<<-1<<endl;
return;
}
continue;
}
if(mm[x]!=2){
cout<<-1<<endl;
return;
}
}
for(auto x:b)
cout<<x<<" ";
cout<<endl;
};