<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 is_palindrome(self, s):
        return s == s[::-1]

    def dfs(self, s, partition = []):
        if s == "":
            self.result.append(partition)
            return

        for i in range(1, len(s) + 1):
            if self.is_palindrome(s[:i]):
                self.dfs(s[i:], partition + [s[:i]])

    def partition(self, s: str) -> List[List[str]]:
        self.result = []
        self.dfs(s)
        return self.result
        
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