<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:
        # def isPal(s):
        #     l,r = 0,len(s)-1
        #     while l < r:
        #         if s[l] != s[r]:
        #             return False
        #         l+=1
        #         r-=1
        #     return True
        isPal = {}
        for i in range(len(s)):
            isPal[(i,i)] = True
        for i in range(len(s)-1):
            isPal[(i,i+1)] = s[i] == s[i+1]
        # print(isPal)
        dp = []
        for i in range(len(s)):
            mcut = float("inf")
            # print(i)
            # print(isPal)
            for j in range(i+1):
                if (j,i) in isPal:
                    ip = isPal[(j,i)]
                else:
                    ip = isPal[(j+1,i-1)] and (s[j]==s[i])
                    isPal[(j,i)]=ip
                if ip:
                    if not j:
                        mcut = 0
                        # break
                    else:
                        mcut = min(mcut, dp[j-1]+1)
            dp.append(mcut)
        return dp[-1]
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