<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>

#include <vector>

using namespace std;

class Solution {
public:
    vector<int> beautifulArray(int n) {
        // Base case: if n is 1, return the trivial array [1]
        if (n == 1) return {1};
        
        vector<int> odd, even;
        
        // Recursive call to generate beautiful array for ceil(n/2)
        vector<int> smallerOdd = beautifulArray((n + 1) / 2);
        vector<int> smallerEven = beautifulArray(n / 2);
        
        // Combine the results for odd and even indices
        for (int x : smallerOdd) odd.push_back(2 * x - 1);
        for (int x : smallerEven) even.push_back(2 * x);
        
        // Combine odd and even parts into the final result
        odd.insert(odd.end(), even.begin(), even.end());
        
        return odd;
    }
};

18000