<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]]:
        ans = []
        def ispalindrome(word):
            return word == word[::-1]

        def backtrack(comb, start): 
            # indicates you have tranversed all letters, and all were eliglble palindromes
            if start == len(s):
                ans.append(comb.copy())
                return
            for i in range(start,len(s)):
                if ispalindrome(s[start:i+1]):
                    comb.append(s[start:i+1])
                    backtrack(comb,i+1)
                    comb.pop()

        backtrack([],0)
        return ans
34108