<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 isPalindrome(self, s):
        length = len(s)
        rev_s = "".join(reversed(s))

        for i in range(length):
            if s[i] != rev_s[i]:
                return False
        return True
        

    def partition(self, s: str) -> List[List[str]]:
        
        sol = []
        def backtrack(index, path):
            if index == len(s):
                res.append(list(path))
            else:
                path.append(s[index])
                backtrack(index+1, path)
                path.pop()

                if path:
                    path[-1] = path[-1] + s[index]
                    backtrack(index+1, path)
    
        res = []
        backtrack(0, [])

        for strL in res:
            pal = True
            for item in strL:
                if not self.isPalindrome(item):
                    pal = False
            if pal == True:
                sol.append(strL)
        
        return sol
        
33293