<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]]:
        partitions = []

        def backtrack(curr_word, curr_partition, i):
            if i == len(s):
                if self.is_palindrome(curr_word):
                    curr_partition.append(curr_word)
                    partitions.append(curr_partition.copy())
                return
            
            curr_word += s[i]
            if self.is_palindrome(curr_word):
                curr_partition.append(curr_word)
                backtrack('', curr_partition.copy(), i + 1)
                curr_partition.pop()
            
            backtrack(curr_word, curr_partition, i + 1)
                

        backtrack('', [], 0)
        return partitions

    def is_palindrome(self, word):
        return word != '' and word == word[::-1]
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