<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 __init__(self):
        self.cache = {}

    def is_palindrome(self, s: str) -> bool:
        N = len(s)
        for i in range(N // 2):
            if s[i] != s[N - 1 - i]:
                return False
        return True

    def partition(self, s: str) -> List[List[str]]:
        if s in self.cache:
            return self.cache[s]

        if len(s) == 1:
            self.cache[s] = [[s]]
            return self.cache[s]

        res = []
        # if s is palindrome
        if self.is_palindrome(s):
            res.append([s])

        for length in range(1, len(s)):
            # separate string
            prefix, suffix = s[:length], s[length:]
            if self.is_palindrome(suffix):
                for prefix_partition in self.partition(prefix):
                    partition = prefix_partition.copy()
                    partition.append(suffix)
                    res.append(partition)
        self.cache[s] = res.copy()
        return res
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