<p>An array <code>nums</code> of length <code>n</code> is <strong>beautiful</strong> if:</p>

<ul>
	<li><code>nums</code> is a permutation of the integers in the range <code>[1, n]</code>.</li>
	<li>For every <code>0 &lt;= i &lt; j &lt; n</code>, there is no index <code>k</code> with <code>i &lt; k &lt; j</code> where <code>2 * nums[k] == nums[i] + nums[j]</code>.</li>
</ul>

<p>Given the integer <code>n</code>, return <em>any <strong>beautiful</strong> array </em><code>nums</code><em> of length </em><code>n</code>. There will be at least one valid answer for the given <code>n</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> n = 4
<strong>Output:</strong> [2,1,4,3]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> n = 5
<strong>Output:</strong> [3,1,2,5,4]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
</ul>

class Solution {
public:
    vector<int> beautifulArray(int n)
    {
        //edge case for if the input is 1
        if (n == 1) 
        {
            return {1};
        }
        
        //variable for holding beautiful array
        vector<int> ba;

        //variables for holding odd and even numbers in the array
        //recursively call the function for the odd and even parts
        vector<int> odd = beautifulArray((n + 1) / 2);  //"odd" part
        vector<int> even = beautifulArray(n / 2);       //"even" part
        
        for (int i : odd)
        {
            ba.push_back(2 * i - 1);  //converting odd indices
        }
        for (int j : even)
        {
            ba.push_back(2 * j);      //converting even indices
        }
        
        return ba;
    }
};
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