<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)
        
        # Step 1: Preprocess the string to determine which substrings are palindromes
        dp = [[False]*n for _ in range(n)]
        for i in range(n):
            dp[i][i] = True
        for i in range(n-1):
            if s[i] == s[i+1]:
                dp[i][i+1] = True
        for l in range(3, n+1):
            for i in range(n-l+1):
                j = i+l-1
                if s[i] == s[j] and dp[i+1][j-1]:
                    dp[i][j] = True
        
        # Step 2: Use dynamic programming to determine the minimum cuts needed
        cuts = list(range(n))
        for i in range(1, n):
            if dp[0][i]:
                cuts[i] = 0
            else:
                for j in range(i):
                    if dp[j+1][i]:
                        cuts[i] = min(cuts[i], cuts[j]+1)
        
        # Step 3: Return the final answer
        return cuts[-1]
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