<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]]:
        self.ans = []
        self.track = []
        self.backtrack(s, 0)
        return self.ans

    def backtrack(self, s, start):
        if start == len(s):
            self.ans.append(copy.deepcopy(self.track))
            return

        for i in range(start, len(s)):
            if not self.is_palindrome(s, start, i):
                continue
            self.track.append(s[start : i + 1])
            self.backtrack(s, i + 1)
            self.track.pop(-1)

    def is_palindrome(self, s, start, end):
        while start < end:
            if s[start] != s[end]:
                return False
            start += 1
            end -= 1
        return True

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