<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]]:
        if s == None or len(s) == 0:
            return []
        self.result = []
        self.backtrack(s, [])
        return self.result
    def backtrack(self, s: str, path: List[int]) -> None:
        #base
        if len(s) == 0:
            self.result.append([num for num in path])
            return
        #logic
        for i in range(len(s)):
            sub = s[:i+1]
            if self.isPalindrome(sub):
                #action
                path.append(sub)
                #recurse
                self.backtrack(s[i+1:], path)
                #backtrack
                path.pop()
    def isPalindrome(self, s:str) -> bool:
        return s == s[::-1]

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