<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, l, r):
        while (l < r):
            if s[l] != s[r]:
                return False
            l += 1
            r -= 1
        return True
    def partition(self, s: str) -> List[List[str]]:
        self.res = []
        def getPartitions(start, currP):

            if start == len(s):

                self.res.append(currP.copy())
                return

            for i in range(start, len(s)):
                if self.isPalindrome(s, start, i):
                    currP.append(s[start:i+1])
                    getPartitions(i + 1, currP)
                    currP.pop()

        getPartitions(0, [])
        return self.res



        
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