<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]]:
        @cache
        def backtrack(s) -> List[List[str]]:
            print(s)
            if len(s) == 0:
                return [[]]

            sol = []
            for i in range(1, len(s) + 1):
                if is_palindrome(s, 0, i - 1):
                    tails = backtrack(s[i:])
                    print(f"{[s[:i]]} + {tails}")
                    for tail in tails:
                        sol.append([s[:i]] + tail)

            return sol

        return backtrack(s)


def is_palindrome(s: str, start: int, end: int) -> bool:
    while start < end:
        if s[start] == s[end]:
            start += 1
            end -= 1
        else:
            return False

    return True

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