<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:
        palindromes = set()
        for i in range(len(s)):
            k = 1
            palindromes.add((i,i))
            while i-k >= 0 and i+k < len(s):
                if s[i-k] == s[i+k]:
                    palindromes.add((i-k,i+k))
                    k += 1
                else:
                    break
            k = 1
            while i-k+1 >= 0 and i+k < len(s):
                if s[i-k+1] == s[i+k]:
                    palindromes.add((i-k+1,i+k))
                    k += 1
                else:
                    break
        
        dp = {-1:-1}
        for i in range(len(s)):
            dp[i] = 1+dp[i-1]
            for k in range(-1, i-1):
                if (k+1,i) in palindromes:
                    dp[i] = min(dp[i], 1+dp[k])
        return dp[len(s)-1]
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