<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]]:
        n = len(s)
        saved = dict()
        if n==1:
            return s
        def isPalindrome(start, end):
            while start <= end:
                if s[start] != s[end]:
                    return False
                start+=1
                end-=1
            return True
        def splits(beg, end):
            res = set()
            if beg == end:
                res.add((s[beg],))
                return res
            if end < beg:
                return res
            curr = s[beg:end+1]
            if curr in saved:
                return saved[curr] 
            if isPalindrome(beg, end):
                res.add((curr,))
            for i in range(end-1, beg-1, -1):
                l_split = splits(beg, i)
                r_split = splits(i+1, end)
                for l in l_split:
                    for r in r_split:
                        res.add(l + r)
            saved[curr] = res
            return res
        return splits(0, n-1)    
        
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