<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):
        if not s or len(s) == 1:
            return True
        return s == s[::-1]
        
    def partition(self, s: str) -> List[List[str]]:
        if not s:
            return []
        if len(s) == 1:
            return [[s]]
        N = len(s)
        DP = [[[],], [[s[0]],]]
        for l in range (2, N+1):
            DP.append([])
            i = l - 1
            for j in range(i,-1,-1):
                cstr = s[j: i+1]
                if self.isPalindrome(cstr):
                    for part in DP[j]:
                        DP[l].append(part + [cstr])
        print(DP)
        return DP[N]
        
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