<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]]:
        # can generate all possible combinations and then check if each is a palindrome
        # checking for a palindrome would be O(N) where n is length of string

        def backtrack(s, idx, cur, ans):
            def isPalindrome(l):
                left = 0
                right = len(l)-1
                while left < right:
                    if l[left] != l[right]:
                        return False
                    left += 1
                    right -= 1
                return True

            if idx == len(s):
                # if is palindrome
                ok = True
                for l in cur:
                    if not isPalindrome(l):
                        ok = False
                        break
                if ok:
                    ans.append(list(cur))
                return

            curString = []
            for i in range(idx, len(s)):
                curString.append(s[i])
                cur.append("".join(curString))
                backtrack(s, i+1, cur, ans)
                cur.pop()

        ans = []
        backtrack(s, 0, [], ans)
        return ans
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