<p>Given an integer <code>n</code>, find a sequence that satisfies all of the following:</p>

<ul>
	<li>The integer <code>1</code> occurs once in the sequence.</li>
	<li>Each integer between <code>2</code> and <code>n</code> occurs twice in the sequence.</li>
	<li>For every integer <code>i</code> between <code>2</code> and <code>n</code>, the <strong>distance</strong> between the two occurrences of <code>i</code> is exactly <code>i</code>.</li>
</ul>

<p>The <strong>distance</strong> between two numbers on the sequence, <code>a[i]</code> and <code>a[j]</code>, is the absolute difference of their indices, <code>|j - i|</code>.</p>

<p>Return <em>the <strong>lexicographically largest</strong> sequence</em><em>. It is guaranteed that under the given constraints, there is always a solution. </em></p>

<p>A sequence <code>a</code> is lexicographically larger than a sequence <code>b</code> (of the same length) if in the first position where <code>a</code> and <code>b</code> differ, sequence <code>a</code> has a number greater than the corresponding number in <code>b</code>. For example, <code>[0,1,9,0]</code> is lexicographically larger than <code>[0,1,5,6]</code> because the first position they differ is at the third number, and <code>9</code> is greater than <code>5</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 3
<strong>Output:</strong> [3,1,2,3,2]
<strong>Explanation:</strong> [2,3,2,1,3] is also a valid sequence, but [3,1,2,3,2] is the lexicographically largest valid sequence.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 5
<strong>Output:</strong> [5,3,1,4,3,5,2,4,2]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 20</code></li>
</ul>

class Solution {
public:
    set<int>mys;
    bool bk(vector<int>&path){
        int p = 0;
        while (p < path.size() && path[p] != -1){p++;}
        if (p == path.size()){return 1;}
        vector<int>cand(mys.begin(), mys.end());
        sort(cand.rbegin(), cand.rend());
        //for (auto& item: path){cout << item << " ";}cout << endl; 
        for (auto& item: cand){
            if (item == 1){path[p] = item; mys.erase(item);}
            else{
                if (p+item >= path.size() || path[p+item] != -1){continue;}
                path[p] = item; path[p+item] = item;
                mys.erase(item);
            }
            if (bk(path)){return 1;}
            else{
                if (item != 1){
                    path[p] = -1; path[p+item] = -1;
                    mys.insert(item);
                }
                else{
                    path[p] = -1; mys.insert(item);
                }

            }
        }
        return 0;
    }
    vector<int> constructDistancedSequence(int n) {
        vector<int>path(2*(n-1)+1, -1);
        for (int i = 1; i <= n; i++){mys.insert(i);}

        //for (int i = n, j= 0; i >= 3; i -= 2, j++){
        //    mys.erase(i);
        //    path[j] = i; path[j+i] = i;
        //}
        bk(path);
        return path;
    }
};
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