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Beautiful Array in C++
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
#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]