<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>

def is_palindrome(s):
    # Remove non-alphanumeric characters and convert to lowercase
    s = ''.join(char.lower() for char in s if char.isalnum())
    
    # Compare the string with its reverse
    return s == s[::-1]


class Solution:
    def partition(self, s):
        print(s)
        self.s = s
        return self.dfs(0, "", [])


    def dfs(self, index, curr_string, curr_list):
        s = self.s
        if index >= len(s):
            return []

        char = s[index]
        curr_string += char

        ret_list = []
        if is_palindrome(curr_string):
            new_list = curr_list.copy()
            new_list.append(curr_string)

            if index == len(s) - 1:
                return [new_list]

            ret_list.extend(self.dfs(index + 1, "", new_list))

        
        ret_list.extend(self.dfs(index + 1, curr_string, curr_list))

        # print(index, curr_string, curr_list, ret_list)

        return ret_list

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