<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]]:  # O(N * 2^N), when all substrings are palindrome, e.g., 'a'*N
        n = len(s)
        @functools.lru_cache(None)
        def dfs(start):
            if start == n: return [[]]
            res = []
            for i in range(start, n):  # O(N)
                cur = s[start:i+1]  # O(N) partition 1
                if cur == cur[::-1]:  # O(N)  # check if it's a palindrome
                    res += [[cur] + rest for rest in dfs(i+1)]  # partition 2
            return res
        return dfs(0)
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