<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 isPal(string):
            print(string)
            #print((string,left,right))
            if len(string)== 1:
                return True
            length = len(string)
            isEven = not((length) %2)
            mid = (length)//2
            R = mid+1
            L = mid-1
            if isEven:
                R = mid
                L = mid-1
            while L >=0 and R < len(string):
                print(L)
                print(R)
                if string[L] != string[R]:
                   # print("False")
                    return False
                L-=1
                R+=1
            return True
       # print(isPal("ab"))
        partition = []
        answer = []
        def backtrack(idx):
            if idx == len(s):
              #  print(partition)
                answer.append(list(partition))           
            for i in range(idx, len(s)):

                if isPal(s[idx:i+1]):
                    partition.append(s[idx:i+1])
                    backtrack(i+1)
                    partition.pop()
            #print(partition)

          #  print(partition)
           # if partition and isPal(partition[-1]):
               # print(partition)
        #answer.append(list(partition))
        backtrack(0)
        return answer




        
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