<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:

        dt = defaultdict(lambda: set())
        n = len(s)
        def exp(i):
            c = 0
            while i-c >= 0 and i+c < n and s[i-c] == s[i+c]:
                dt[i-c].add(i+c)
                c += 1
            c = 0
            while i-c >= 0 and i+c+1 < n and s[i-c] == s[i+c+1]:
                dt[i-c].add(i+c+1)
                c += 1
        for x in range(n): exp(x)
        
   
        @lru_cache(None)
        def dfs(i):
            if i == n: return 0
            return 1 + min([dfs(x+1) for x in dt[i]])
        return dfs(0)-1
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