<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:
        def per(s,k=0,n=0):
            if len(s)<2:
                return s
            p=[]
            for i in range(len(s)):
                x=s[i]
                xl=s[:i]+s[i+1:]
                for b in per(xl):
                    if len(s)==n and len(p)==k:
                        break
                    p.append(x+b)
                if len(s)==n and len(p)==k:
                    break
            if len(s)==n and len(p)==k:
                return p[-1]
            return p
        s="01234567890"
        s=s[1:n+1]
        return per(s,k,n)

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