<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]]:
        result = []

        def palindrome(s: str) -> bool:
            return s == s[::-1] # compare to its reverse
        
        # s = the string
        # j = current starting index for substring exploration
        # path = current partition (to add to result)
        # result = all complete paths
        def dfs(s: str, j: int, path: List[str], result: List[List[str]]) -> None:
            if j == len(s):
                result.append(path)
                return
            for i in range(j, len(s)):
                if palindrome(s[j: i + 1]):
                    dfs(s, i + 1, path + [s[j: i + 1]], result)
            
        dfs(s, 0, [], result)
        return result
        
    # Time: O(n*2^n)
    # Space: O(n)
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