<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:
        dic={}
        n=len(s)
        palindromeTable = [[False] * n for _ in range(n)]
        for i in range(n):
            palindromeTable[i][i] = True
        for prefix_len in range(2, n + 1):
            for start in range(n - prefix_len + 1):
                end = start + prefix_len - 1
                if s[start] == s[end]:
                    if prefix_len == 2 or palindromeTable[start + 1][end - 1]:
                        palindromeTable[start][end] = True

        def partition(start:int,end:int) -> int:
            nonlocal dic
            if start>end:
                return []
            if (start,end) in dic:
                return dic[(start,end)]
            res=end-start+1
            for i in range(start,end+1):
                if palindromeTable[start][i]:
                    if i==end:
                        res=0
                    else:
                        res=min(res,partition(i+1,end)+1)
            dic[(start,end)]=res
            return res
        return partition(0,len(s)-1)
        
94773