<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) {
    vector<int> A(n);
    iota(A.begin(), A.end(), 1);
    divide(A, 0, n - 1, 1);
    return A;
  }

 private:
  void divide(vector<int>& A, int l, int r, int mask) {
    if (l >= r)
      return;
    const int m = partition(A, l, r, mask);
    divide(A, l, m, mask << 1);
    divide(A, m + 1, r, mask << 1);
  }

  int partition(vector<int>& A, int l, int r, int mask) {
    int nextSwapped = l;
    for (int i = l; i <= r; ++i)
      if (A[i] & mask)
        swap(A[i], A[nextSwapped++]);
    return nextSwapped - 1;
  }
};
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