<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)

        palindromes = set()
        # precompute all palindromes in n^2
        for i in range(len(s)):
            # odd
            l = r = i
            while l >= 0 and r < N and s[l] == s[r]:
                palindromes.add((l, r))
                l -= 1
                r += 1
            
            # even
            l, r = i, i + 1
            while l >= 0 and r < N and s[l] == s[r]:
                palindromes.add((l, r))
                l -= 1
                r += 1

        
        cache = {}
        def dfs(index):
            if index >= N:
                return -1
            
            if index in cache:
                return cache[index]
            
            res = float('inf')
            for end in range(index, N):
                if (index, end) in palindromes:
                    res = min(res, 1 + dfs(end + 1))
            
            cache[index] = res
            return res
        
        return dfs(0)

        
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