Construct the Lexicographically Largest Valid Sequence
<p>Given an integer <code>n</code>, find a sequence with elements in the range <code>[1, n]</code> 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>
Array; Backtracking
class Solution {
public:
    int n;
    vector<int> cnt, path;

    vector<int> constructDistancedSequence(int _n) {
        n = _n;
        cnt.resize(n * 2, 2);
        path.resize(n * 2);
        cnt[1] = 1;
        dfs(1);
        vector<int> ans;
        for (int i = 1; i < n * 2; ++i) ans.push_back(path[i]);
        return ans;
    }

    bool dfs(int u) {
        if (u == n * 2) return 1;
        if (path[u]) return dfs(u + 1);
        for (int i = n; i > 1; --i) {
            if (cnt[i] && u + i < n * 2 && !path[u + i]) {
                path[u] = path[u + i] = i;
                cnt[i] = 0;
                if (dfs(u + 1)) return 1;
                cnt[i] = 2;
                path[u] = path[u + i] = 0;
            }
        }
        if (cnt[1]) {
            path[u] = 1;
            cnt[1] = 0;
            if (dfs(u + 1)) return 1;
            cnt[1] = 1;
            path[u] = 0;
        }
        return 0;
    }
};