<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]]:
        output = []
        def is_palindrome(s):
            s = s.replace(" ", "").lower()
            # Check if the string is equal to its reverse
            return s == s[::-1]

        def dfs(string):
            if len(string)==1:
                return [[string]]
            stores = []
            for i in range(len(string)):
                firstString = string[0:i+1]
                if is_palindrome(firstString):
                    if i+1 == len(string):
                        stores.append([firstString])
                        return stores
                    secondString = dfs(string[i+1:])
                    for store in secondString:
                        store.insert(0,firstString)
                    stores.extend(secondString)
            return stores
            
        return dfs(s)
                    


    

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