<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:
    def makeBeautiful(self, arr):
        if len(arr) < 3:
            return arr

        odds = []
        evens = []

        for n in arr:
            if n % 2 == 1:
                odds.append(n)
            else:
                evens.append(n)

        if len(odds) > 2:
            odds = [e*2+1 for e in self.makeBeautiful([(e-1)//2 for e in odds])]

        if len(evens) > 2:
            evens = [e*2 for e in self.makeBeautiful([e//2 for e in evens])]

        return odds+evens

    def beautifulArray(self, n: int) -> List[int]:
        return self.makeBeautiful([i for i in range(1, n+1)])
        
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