1385B - Restore the Permutation by Merger.cpp

/*
A permutation of length n is a sequence of integers from 1 to n of length n containing each number exactly once. For example, [1], [4,3,5,1,2], [3,2,1] are permutations, and [1,1], [0,1], [2,2,1,4] are not.

There was a permutation p[1…n]. It was merged with itself. In other words, let's take two instances of p and insert elements of the second p into the first maintaining relative order of elements. The result is a sequence of the length 2n.

For example, if p=[3,1,2] some possible results are: [3,1,2,3,1,2], [3,3,1,1,2,2], [3,1,3,1,2,2]. The following sequences are not possible results of a merging: [1,3,2,1,2,3], [3,1,2,3,2,1], [3,3,1,2,2,1].

For example, if p=[2,1] the possible results are: [2,2,1,1], [2,1,2,1]. The following sequences are not possible results of a merging: [1,1,2,2], [2,1,1,2], [1,2,2,1].

Your task is to restore the permutation p by the given resulting sequence a. It is guaranteed that the answer exists and is unique.

You have to answer t independent test cases.

Input
The first line of the input contains one integer t (1=t=400) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (1=n=50) — the length of permutation. The second line of the test case contains 2n integers a1,a2,…,a2n (1=ai=n), where ai is the i-th element of a. It is guaranteed that the array a represents the result of merging of some permutation p with the same permutation p.

Output
For each test case, print the answer: n integers p1,p2,…,pn (1=pi=n), representing the initial permutation. It is guaranteed that the answer exists and is unique.

Example
inputCopy
5
2
1 1 2 2
4
1 3 1 4 3 4 2 2
5
1 2 1 2 3 4 3 5 4 5
3
1 2 3 1 2 3
4
2 3 2 4 1 3 4 1
outputCopy
1 2 
1 3 4 2 
1 2 3 4 5 
1 2 3 
2 3 4 1 
*/



#include<bits/stdc++.h>
using namespace std;

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    
    int t;
    cin >> t;
    
    while (t--) {
	    int n;
	    cin >> n;
	    
	   unordered_set <int> s;
	   
	   for (int i = 0; i < 2 * n; i++) {
	   		int a;
	   		cin >> a;
	   		
	   		if (s.find(a) == s.end()) {
	   			s.insert(a);
	   		    cout << a << " ";
			   }
	   }
	   
	   cout << endl;
	}
  	
    return 0;
}