<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)
        isPalindrome=[[0]*n for i in range(n)]
        for l in range(1,n+1):
            for i in range(n-l+1):
                j=i+l-1
                if i==j:
                    isPalindrome[i][j]=True
                elif s[i]==s[j] and (j==i+1 or isPalindrome[i+1][j-1]):
                    isPalindrome[i][j]=True
                else:
                    isPalindrome[i][j]=False
        dp=[[None]*n for i in range(n)] #memoization table 
        def Partition(start,end):
            #base case
            if start==end or isPalindrome[start][end]:
                return 0
            #recurrsive case
            if dp[start][end] is not None:
                return dp[start][end]
            minimum=end-start
            for end_idx in range(start,end):
                if isPalindrome[start][end_idx]:
                    minimum=min(minimum,1+Partition(end_idx+1,end))
            dp[start][end]=minimum
            return dp[start][end]
            


        return Partition(0,n-1)

            
                
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