<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]]:
        slen = len(s)

        def isPalin(start,end):
            l = start
            r = end
            while(l<r):
                if s[l]!=s[r]:
                    return False
                l+=1
                r-=1
            return True

        memo = dict()

        def f(start):
            toReturn = []
            if start == slen:
                return []
            
            if start in memo:
                return memo[start]
            
            subString = ""
            for i in range(start,slen):
                subString += s[i]
                
                if isPalin(start,i):
                    
                    nexts = f(i+1)
                    if not nexts:
                        toReturn.append([subString])
                    else:
                        for next in nexts:
                            toReturn.append([subString]+next)
            
            memo[start] = toReturn[:]
            return toReturn
        
        return f(0)
29218