<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>

import functools

class Solution:
    def minCut(self, s: str) -> int:
        string = s
        @functools.cache
        def isPalindrome(s, e):
            while s<e:
                if string[s] != string[e]:
                    return False
                s += 1
                e -= 1
            else: return True

        @functools.cache
        def recur(currindex):
            if currindex >= len(string) or isPalindrome(currindex, len(string) -1):
                return 0
            sol = math.inf
            for i in range(len(string)-1, currindex - 1, -1):
                if isPalindrome(currindex, i):
                    sol = min(sol, recur(i + 1) + 1)
                    if sol == 0 or sol == 1:
                        return sol
            return sol

        return recur(0)
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