<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):
        otherComb = self.helper(s)
        ret = []

        for i in otherComb:
            flag = True
            for word in i:
                if(word != word[::-1]):
                    flag = False
                    break
            if(flag):
                ret.append(i)

        return ret

    def helper(self, s):
        ret = []
        
        # Base Case
        if(len(s) == 1):
            return [[s]]
        
        current = s[0]
        s = s[1:]

        otherComb = self.helper(s)
        print(s, otherComb)
        # For each combo add 2 first with concat other append
        for i in otherComb:
            if(i[-1] == i[-1][::-1]):
                ret.append([current] + i)
            ret.append([current + i[0]] + i[1:])

        return ret
        
        
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