<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: int = len(s)
        if n == 1 or len(set(s)) == 1:
            return 0
        if n == 2:
            return 0 if s[0] == s[1] else 1

        @cache
        def dfs(i: int, j: int) -> int:
            if s[i:j] == s[i:j][::-1]:
                return 0

            result: int = float('inf')
            for k in range(i + 1, j):
                if s[i:k] == s[i:k][::-1]:
                    result = min(result, dfs(i, k) + dfs(k, j) + 1)
            return result

        return dfs(0, n)

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