<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 backtrack(self, start, path):
        if start == len(self.s):
            self.ans.append(path[:])
            return

        for end in range(start + 1, len(self.s) + 1):
            sub = self.s[start : end]

            if sub == sub[::-1]:
                self.backtrack(end, path + [sub])

    def partition(self, s: str) -> List[List[str]]:
        '''
        find a palindrome, backtrack to find others in remainder of string
        '''
        self.ans = []
        self.s = s
        self.backtrack(0, [])
        return self.ans

    
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