<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 validPartitioner(self, i, combination):
        # exceeded str
        if i == self.n:
            self.ans.add(tuple(combination))
            return

        for j in range(i, self.n):
            if self.dp[i][j] == 1:
                subStr = self.s[i:(j+1)]
                combination.append(subStr)
                self.validPartitioner(j+1, combination)
                combination.pop()

    def partition(self, s: str) -> List[List[str]]:
        self.n, self.s = len(s), s

        self.dp = [[-1]*self.n for _ in range(self.n)]
        for i in range(self.n):
            self.dp[i][i] = 1 # every single char is a palindrome
            
        for length in range(2, self.n+1):
            for i in range(self.n-length+1):

                if length == 2:
                    self.dp[i][i+length-1] = 1 if self.s[i] == self.s[i+length-1] else 0
                else:
                    self.dp[i][i+length-1] = 1 if (self.s[i] == self.s[i+length-1] and self.dp[i+1][i+length-2] == 1) else 0

        self.ans = set()
        self.validPartitioner(0, [])
        return [list(c) for c in self.ans]
        
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