# 0932-beautiful-array


Try it on <a href='https://leetcode.com/problems/0932-beautiful-array'>leetcode</a>

## Description
<div class="description">
<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>

</div>

## Solution(Python)
```Python
from functools import lru_cache


class Solution:
    def beautifulArray(self, N):
        @lru_cache
        def f(N): # 2
            if N <= 1:
                return [1]
            odds = f((N+1)//2) # 
            evens = f(N//2)
            ans = [2*x-1 for x in odds] + [2*x for x in evens]
            return ans
        return f(N)
```