<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]]:
        current, res, visited = [], set(), set()
        self.backtrack(s, current, res, 0, visited)
        return list(res)

    def backtrack(self, s, current, res, i, visited):
        if (tuple(current), i) in visited:
            return
        if i == len(s):
            allPalindrome = True
            for sub in current:
                if sub != sub[::-1]:
                    allPalindrome = False
            if allPalindrome:
                res.add(tuple(current))
            return
        for j in range(i + 1, len(s) + 1):
            # i=0, current = "a", j = 1
                # i=1, current ["a", "a"], j =2
                    # i=2, current ["a", "a", "b"], j =3
                # i=1, current ["a", "ab"], j =3

            # i=0, current = "aa", j = 2
                # i=2, current ["aa", "b"], j = 3

            # i=0, current = "aab", j = 3 -> allPalindrome = false => return
            if (tuple(current + [s[i:j]]), i) not in visited:
                visited.add((tuple(current + [s[i:j]]), i))
                self.backtrack(s, current + [s[i:j]], res, j, visited)
                visited.remove((tuple(current + [s[i:j]]), i))

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