<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]]:
        ss = '#' + '#'.join(s) + '#'
        n, m = len(s), len(ss)
        dp = [[list() for __ in range(n)] for _ in range(n)]
        for i in range(m):
            j = 0
            while i - j > 0 and i + j < m:
                if ss[i-j] != '#': 
                    if ss[i-j] == ss[i+j]:
                        a, b = (i-j-1) // 2, (i+j-1) // 2
                        dp[a][b].append([s[a : b+1]])

                    else:
                        break
                j += 1
        
        # for r in dp:
        #     print(r)
        
        for j in range(1, n):
            for i in range(1, j+1):
                if dp[i][j]:
                    for x in dp[0][i-1]:
                        dp[0][j].append(x + dp[i][j][0])
        
            
        
        # for r in dp:
        #     print(r)


        return dp[0][-1]
26365