<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 is_palindrome(self, s: str) -> bool:
        l, r = 0, len(s) - 1
        while l < r:
            if s[l] != s[r]:
                return False
            l += 1
            r -= 1
        return True

    def partition(self, s: str) -> list[list[str]]:
        res = []

        def backtrack(s2: str, curr: list[str]):
            if len(s2) < 1:
                res.append(curr)
                return
            for i in range(len(s2)):
                sub = s2[:i + 1]
                if self.is_palindrome(sub):
                    new = curr.copy()
                    new.append(sub)
                    backtrack(s2[i + 1:], new)

        backtrack(s, [])

        return res

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