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:
    def constructDistancedSequence(self, n: int) -> List[int]:
        def dfs(u):
            if u == n * 2:
                return True
            if path[u]:
                return dfs(u + 1)
            for i in range(n, 1, -1):
                if cnt[i] and u + i < n * 2 and path[u + i] == 0:
                    cnt[i] = 0
                    path[u] = path[u + i] = i
                    if dfs(u + 1):
                        return True
                    path[u] = path[u + i] = 0
                    cnt[i] = 2
            if cnt[1]:
                cnt[1], path[u] = 0, 1
                if dfs(u + 1):
                    return True
                path[u], cnt[1] = 0, 1
            return False

        path = [0] * (n * 2)
        cnt = [2] * (n * 2)
        cnt[1] = 1
        dfs(1)
        return path[1:]
