<p>Given a string <code>s</code>, partition <code>s</code> such that every <span data-keyword="substring-nonempty">substring</span> of the partition is a <span data-keyword="palindrome-string"><strong>palindrome</strong></span>. Return <em>all possible palindrome partitioning of </em><code>s</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> s = "aab"
<strong>Output:</strong> [["a","a","b"],["aa","b"]]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> s = "a"
<strong>Output:</strong> [["a"]]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 16</code></li>
	<li><code>s</code> contains only lowercase English letters.</li>
</ul>

class Solution:
    def partition(self, s: str) -> List[List[str]]:
        def f(s,l,a,p):
            def g(s,l,r):
                while l<r:
                    if s[l]!=s[r]:return False
                    l,r=l+1,r-1
                return True
            if a==l:yield p;return;
            for b in range(a,l):
                if g(s,a,b):
                    p.append(s[a:b+1])
                    yield from f(s,l,b+1,p)
                    p.pop()
        yield from f(s,len(s),0,deque())

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