<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]]:
        len_s = len(s)
        dp = [[False] * len_s for _ in range(len_s)]
        result = []
        self.dfs(result, s, 0, [], dp)
        return result

    def dfs(
        self,
        result: List[List[str]],
        s: str,
        start: int,
        currentList: List[str],
        dp: List[List[bool]],
    ):
        if start >= len(s):
            result.append(list(currentList))
        for end in range(start, len(s)):
            if s[start] == s[end] and (
                end - start <= 2 or dp[start + 1][end - 1]
            ):
                dp[start][end] = True
                currentList.append(s[start : end + 1])
                self.dfs(result, s, end + 1, currentList, dp)
                currentList.pop()
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