<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]]:
        max_mask = 2**(len(s) - 1)
        ans = []
        for mask in range(max_mask):
            new_partition = []
            flag = True
            temp = ""
            old = (mask & 1)
            for i in range(len(s)):
                if (mask >> i & 1) == old:
                    temp += s[i]
                else:
                    old = (mask >> i & 1)
                    # print(temp)
                    if not self.palindrome(temp):
                        flag = False
                        break
                    new_partition.append(temp)
                    temp = s[i]
            if flag:
                if self.palindrome(temp):
                    new_partition.append(temp)
                    ans.append(new_partition)
        return ans

    def palindrome(self, s):
        l = 0
        r = len(s) -1
        while l < r:
            if s[l] != s[r]:
                return False
            l += 1
            r -= 1
        return True
38998