<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]]:
        dp = []
        for i in range(len(s)):
            dp.append([s[i]])
            left, right = i - 1, i + 1
            while left >= 0 and right < len(s):
                if s[left] == s[right]:
                    dp[left].append(s[left:right+1])
                    left -= 1
                    right += 1
                else:
                    break
            if i < len(s) - 1 and s[i] == s[i+1]:
                dp[-1].append(s[i:i+2])
                left, right = i-1, i+2
                while left >= 0 and right < len(s):
                    if s[left] == s[right]:
                        dp[left].append(s[left:right+1])
                        left -= 1
                        right += 1
                    else:
                        break
        def compute(length):
            if length == len(s):
                return [[]]
            res = []
            for w in dp[length]:
                for nxt in compute(length + len(w)):
                    res.append([w] + nxt)
            return res
        return compute(0)

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