# Beautiful Array in C++

C++Server Side ProgrammingProgramming

Suppose for some fixed value of N, an array A is beautiful when it is a permutation of the integers 1, 2, ..., N, such that −

• For every i < j, there is no such k with i < k < j such that A[k] * 2 = A[i] + A[j].

Suppose we have N, we have to find any beautiful array A.

So if the input is like 5, then the output will be [3,1,2,5,4]

To solve this, we will follow these steps −

• Create one array called ret, insert 1 into ret

• while size of ret < N

• create an array temp

• for i in range 0 to size of ret – 1

• if ret[i] * 2 – 1 <= N, then insert ret[i] * 2 – 1 into temp array

• for i in range 0 to size of ret – 1

• if ret[i] * 2 <= N, then insert ret[i] * 2 into temp array

• set ret := temp

• return ret

Let us see the following implementation to get better understanding −

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<auto> v){
cout << "[";
for(int i = 0; i<v.size(); i++){
cout << v[i] << ", ";
}
cout << "]"<<endl;
}
class Solution {
public:
vector<int> beautifulArray(int N) {
vector <int> ret;
ret.push_back(1);
while(ret.size() < N){
vector <int> temp;
for(int i = 0; i < ret.size(); i++){
if(ret[i] * 2 - 1 <= N) temp.push_back(ret[i] * 2 - 1);
}
for(int i = 0; i < ret.size(); i++){
if(ret[i] * 2 <= N)temp.push_back(ret[i] * 2 );
}
ret = temp;
}
return ret;
}
};
main(){
Solution ob;
print_vector(ob.beautifulArray(6));
}

### Input

5

## Output

[1,5,3,2,4]
Published on 10-Apr-2020 13:41:44