<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)
        dp = [[(s[i] == s[j]) for j in range(n)] for i in range(n)]

        # build dp differently 
        for diag in range(2, n):
            for i in range(n-diag):
                if not dp[i+1][i+diag-1]:
                    dp[i][i+diag] = False

        answers = []

        def backtrack(output, idx):
            if idx == n:
                answers.append([string for string in output])
                return
            
            i = idx
            for j in range(i, n):
                if (i == j or \
                    (j == i+1 and dp[i][j-1]) or \
                    (j > i+1 and i < n-1 and dp[i+1][j-1])) \
                        and dp[i][j]:

                    output.append(s[i:j+1])
                    backtrack(output, j+1)
                    output.pop()

        backtrack([], 0)
        return answers
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