<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">palindrome</span>.</p>

<p>Return <em>the <strong>minimum</strong> cuts needed for a palindrome partitioning of</em> <code>s</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;aab&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> The palindrome partitioning [&quot;aa&quot;,&quot;b&quot;] could be produced using 1 cut.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;a&quot;
<strong>Output:</strong> 0
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;ab&quot;
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 2000</code></li>
	<li><code>s</code> consists of lowercase English letters only.</li>
</ul>

class Solution:
    def minCut(self, s: str) -> int:
        memoCuts = [[None] * len(s) for _ in range(len(s))]
        memoPalindrome = [[None] * len(s) for _ in range(len(s))]

        @cache
        def isPalindrome(start, end):
            if start >= end:
                return True
            return s[start:end+1] == s[start:end+1][::-1]

        @cache
        def findMinimumCut(start, end):
            if start == end or isPalindrome(start, end):
                return 0
            minimumCut = math.inf
            for i in range(start, end + 1):
                if isPalindrome(start, i):
                    minimumCut = min(minimumCut, 1 + findMinimumCut(i + 1, end))
            return minimumCut

        return findMinimumCut(0, len(s) - 1)

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