<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]]:
        ans = []

        def dfs(i: int, curr: str):
            if i == len(s):
                ans.append(curr)

            for j in range(i, len(s)):
                sol = s[i : j + 1]
                if self.is_palindrome(sol):
                    dfs(j + 1, [*curr, sol])

        dfs(0, [])
        return ans

    def is_palindrome(self, s: str) -> bool:
        n = len(s)
        return all(s[i] == s[n - i - 1] for i in range(n // 2))

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