<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]]:
        path = []
        result = []

        len_s = len(s)

        def is_Palindrome(s, start_idx, end_idx):
            s_in = s[start_idx:end_idx+1]
            return s_in == s_in[::-1]

        def backtracking(start_idx):
            if start_idx >= len_s:
                result.append(path.copy())

            for i in range(start_idx, len_s):
                if is_Palindrome(s, start_idx, i):
                    path.append(s[start_idx:i + 1])
                    backtracking(i + 1)
                    path.pop()

        backtracking(0)
        return result
33904