<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 isPalindrome(self, s):
        return s == s[::-1]
    
    def partition(self, s: str, cache = None) -> List[List[str]]:
        if not s:
            return [[]]
        if not cache:
            cache = dict()
        if len(s) == 1:
            cache[s] = [[s]]
        if s in cache:
            return cache[s]
        else:
            for i in range(1,len(s)+1):
                if self.isPalindrome(s[:i]):
                    substringPartitions = self.partition(s[i:],cache)
                    res = []
                    if substringPartitions:
                        for substringPartition in substringPartitions:
                            res.append([s[:i]]+substringPartition)
                        if s in cache:
                            cache[s] = cache[s] + res
                        else:
                            cache[s] = res
            return cache[s] if s in cache else []
                    
                    
        
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