<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]]:
        N = len(s)
        if N == 0:
            return []
        is_pal = [[False] * N for i in range(N)]
        for i in range(N):
            is_pal[i][i] = True

        for l in range(2, N + 1):
            for i in range(N):
                j = i + l - 1
                if j >= N:
                    break
                if l == 2:
                    is_pal[i][j] = s[i] == s[j]
                else:
                    is_pal[i][j] = (s[i] == s[j]) and is_pal[i + 1][j - 1]
        
        ans = [[[]]]
        for i in range(0, N):
            result = []
            for j in range(0, i + 1):
                if is_pal[j][i]:
                    sub = s[j: i + 1]
                    prev = ans[j - 1 + 1]
                    for x in prev:
                        result.append(x + [sub])  
            ans.append(result)

        return ans[-1]



        
26161