<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>
#include <unordered_map>

class Solution {
public:
    std::unordered_map<int, std::vector<int>> memo;

    std::vector<int> beautifulArray(int N) {
        return f(N);
    }

private:
    std::vector<int> f(int N) {
        if (memo.find(N) != memo.end()) {
            return memo[N];
        }

        std::vector<int> ans(N);
        if (N == 1) {
            ans[0] = 1;
        } else {
            int t = 0;
            for (int x : f((N + 1) / 2)) { // odds
                ans[t++] = 2 * x - 1;
            }
            for (int x : f(N / 2)) { // evens
                ans[t++] = 2 * x;
            }
        }
        memo[N] = ans;
        return ans;
    }
};

9700