<p>The set <code>[1, 2, 3, ...,&nbsp;n]</code> contains a total of <code>n!</code> unique permutations.</p>

<p>By listing and labeling all of the permutations in order, we get the following sequence for <code>n = 3</code>:</p>

<ol>
	<li><code>&quot;123&quot;</code></li>
	<li><code>&quot;132&quot;</code></li>
	<li><code>&quot;213&quot;</code></li>
	<li><code>&quot;231&quot;</code></li>
	<li><code>&quot;312&quot;</code></li>
	<li><code>&quot;321&quot;</code></li>
</ol>

<p>Given <code>n</code> and <code>k</code>, return the <code>k<sup>th</sup></code> permutation sequence.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> n = 3, k = 3
<strong>Output:</strong> "213"
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> n = 4, k = 9
<strong>Output:</strong> "2314"
</pre><p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> n = 3, k = 1
<strong>Output:</strong> "123"
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 9</code></li>
	<li><code>1 &lt;= k &lt;= n!</code></li>
</ul>

class Solution:
    def getPermutation(self, n: int, k: int) -> str:
        path = []
        pathSet = set()
        res = []
        space = [i for i in range(1, n + 1)]
        def backtrack() -> bool:
            if len(pathSet) == n:
                # print(path)
                res.append(path)
                if len(res) == k:
                    return True
                return False
            
            for i in range(len(space)):
                if space[i] not in pathSet:
                    path.append(space[i])
                    pathSet.add(space[i])

                    if backtrack():
                        return True
                    
                    path.pop()
                    pathSet.remove(space[i])

            return False
        
        backtrack()
        return "".join([str(x) for x in res[-1]])
45300