<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 backtracking(self, s, index, path, res):
        if index == len(s):
            res.append(path[:])
            return
        for i in range(index, len(s)):
            if s[index:i+1] == s[index:i+1][::-1]:
                path.append(s[index:i+1])
                self.backtracking(s, i+1, path, res)
                path.pop()

    def partition(self, s: str) -> List[List[str]]:

        res = []
        self.backtracking(s, 0, [], res)
        return res


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