<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>

mem=[-1 for x in range(2000)]
class Solution:
    def minCut(self, s: str) -> int:  
        ps=set()
        def chp(i,j):
            if i==j:return 1
            return (i,j) in ps
        for c in range(len(s)):
            mem[c]=-1
            i=c
            j=c
            while i>=0 and j<len(s):
                if s[i]!=s[j]:break
                ps.add((i,j))
                i-=1
                j+=1
            i=c
            j=c+1
            while i>=0 and j<len(s):
                if s[i]!=s[j]:break
                ps.add((i,j))
                i-=1
                j+=1
                
        
        def dp(i):
            if i>=len(s):return 0
            if mem[i]!=-1:return mem[i]
            if chp(i,len(s)-1):return 0
            a=1<<30
            for nx in range(i+1,len(s)):
                if not chp(i,nx-1):continue
                a=min(a,1+dp(nx))
            mem[i]=a
            return a
        return dp(0)
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