<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]]:
        length = len(s)
        def helper(l, r): 
            while l >= 0 and r < length and s[l] == s[r]:
                palindromes[l].append(s[l:r+1]) 
                l -= 1 
                r += 1
                

        palindromes = defaultdict(list) 
        for i in range(length): 
            if i != length - 1: 
                helper(i, i+1)
            helper(i, i) 
        
        d = deque([]) 
        res = [] 
        for pali in palindromes[0]: 
            d.append([pali]) 

        while d: 
            path = d.popleft()
            curr = ''.join(path)
            length = len(curr) 
            if length == len(s): 
                res.append(path) 

            for pali in palindromes[length]: 
                d.append(path + [pali])  # Append a new path with the palindrome added

        return res

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