<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]]:
        res = []
        partition = []
        
        def isPalindrome(s, l, r): 
            while l < r: 
                if s[l] != s[r]: 
                    return False 
                l += 1
                r -= 1
            return True 

        def backtrack(i): 
            if i >= len(s):
                res.append(partition[:]) 
                return 
            for j in range(i, len(s)): 
                if isPalindrome(s, i, j): 
                    partition.append(s[i: j + 1]) 
                    backtrack(j + 1)
                    partition.pop() 
        backtrack(0) 
        return res


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