<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]]:
        def isPalindrome(string, start, end):
            substr = string[start: end + 1]
            return substr == substr[::-1]
        
        def palindromePartitioning(string, ind, subStr, ans):
            if ind == len(string):
                ans.append(subStr[:])
                return 
                
                
            for i in range(ind, len(string)):
                if isPalindrome(string, ind, i):
                    subStr.append(string[ind : i + 1])
                    palindromePartitioning(string, i + 1, subStr, ans)
                    subStr.pop()
                    
                    
                    
        # string = "aaff"
        ind = 0
        subStr = []
        ans = []
        # print(palindromePartitioning(s, ind, subStr, ans))
        palindromePartitioning(s, ind, subStr, ans)
        return ans
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