<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]]:
        result_map: dict[str, list[list[str]]] = {}

        def is_palindrome(s: str) -> bool:
            left, right = 0, len(s) - 1
            while left < right:
                if s[left] != s[right]:
                    return False
                left, right = left + 1, right - 1

            return True

        def helper(s: str) -> list[list[str]]:
            if len(s) == 0:
                return []

            nonlocal result_map
            if s in result_map:
                return result_map[s]

            result: list[list[str]] = []
            for i in range(1, len(s)):
                if is_palindrome(s[:i]):
                    for r in helper(s[i:]):
                        result.append([s[:i]] + r)

            if is_palindrome(s):
                result.append([s])

            result_map[s] = result
            return result

        return helper(s)        
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