<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:
        self.res = ""
        self.count = 0
        self.stop = False
        self.backtrack(n, k, "", [False] * n)
        print(self.stop)
        return self.res
    
    def backtrack(self, n, k, perm, used):
        if self.stop:
            return

        if len(perm) == n:
            self.count += 1
            if self.count == k:
                self.res = perm
                self.stop = True
            return

        for i in range(n):
            if not used[i]:
                used[i] = True
                self.backtrack(n, k, perm + str(i+1), used)
                used[i] = False
        


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