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
func constructDistancedSequence(n int) []int {
	path := make([]int, n*2)
	cnt := make([]int, n*2)
	for i := range cnt {
		cnt[i] = 2
	}
	cnt[1] = 1
	var dfs func(u int) bool
	dfs = func(u int) bool {
		if u == n*2 {
			return true
		}
		if path[u] > 0 {
			return dfs(u + 1)
		}
		for i := n; i > 1; i-- {
			if cnt[i] > 0 && u+i < n*2 && path[u+i] == 0 {
				cnt[i] = 0
				path[u], path[u+i] = i, i
				if dfs(u + 1) {
					return true
				}
				cnt[i] = 2
				path[u], path[u+i] = 0, 0
			}
		}
		if cnt[1] > 0 {
			cnt[1] = 0
			path[u] = 1
			if dfs(u + 1) {
				return true
			}
			cnt[1] = 1
			path[u] = 0
		}
		return false
	}
	dfs(1)
	return path[1:]
}