<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:
        n = len(s)
        dp = [[False] * n for _ in range(n)]

        for odd_center in range(n):
            c = 0
            while 0 <= odd_center - c and odd_center + c <= n - 1 and s[odd_center - c] == s[odd_center + c]:
                dp[odd_center - c][odd_center + c] = True
                c += 1
        for even_center in range(n - 1):
            c = 0
            while 0 <= even_center - c and even_center + c + 1 <= n - 1 and s[even_center - c] == s[even_center + c + 1]:
                dp[even_center - c][even_center + c + 1] = True
                c += 1
        
        # gets number of palindromes required to maximally palindromize substring s[i:j+1]
        @cache
        def recur(i, j) -> int:
            if i > j:
                return 0
            else:
                ans = math.inf
                substring_length = j - i + 1
                for l in range(substring_length, 0, -1):
                    if dp[i][i + l - 1]:
                        ans = min(ans, 1 + recur(i + l, j))
                return ans

        return recur(0, n - 1) - 1 # minus 1 because number of parts = number of cuts + 1

        
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