<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]]:
        def isPalindrome(s):
            return s == s[::-1]

        def backtracking(s, curr):
            if s == "":
                return [curr]
            ans = []
            for i in range(1, len(s) + 1):
                if isPalindrome(s[:i]):
                    res = backtracking(s[i:], curr + [s[:i]])
                    if res:
                        ans += res
            return ans if ans else None

        
        return backtracking(s, [])


        ### Neetcode Backtracking
        # res, part = [], []
        # def isPali(s, l, r):
        #     while l < r:
        #         if s[l] != s[r]:
        #             return False
        #         l, r = l + 1, r - 1
        #     return True
        # def dfs(i):
        #     if i >= len(s):
        #         res.append(part.copy())
        #         return
        #     for j in range(i, len(s)):
        #         if isPali(s, i, j):
        #             part.append(s[i : j + 1])
        #             dfs(j + 1)
        #             part.pop()
        # dfs(0)
        # return res

    
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