<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:
        
        # Compute all palindromes
        is_palindrome={}
        for i in range(1,len(s)+1):
            for j in range(0,len(s)+1-i):
                if i==1:
                    is_palindrome[(j,j+i)]=True
                elif i==2:
                    is_palindrome[(j,j+i)]=s[j]==s[j+1]
                else:
                    curr_matches=s[j]==s[j+i-1]
                    is_palindrome[(j,j+i)]=curr_matches and is_palindrome[(j+1,j+i-1)]
            
        dp = [0 for i in range(len(s)+1)]

        max_cuts = len(s)-1


        for i in range(1,len(s)+1):
            curr_min=max_cuts

            # no cut needed
            if is_palindrome[(0,i)]:
                dp[i-1] = 0
                continue
            for j in range(1,i):
                if is_palindrome[(j,i)]:
                    old_min=curr_min
                    curr_min = min(curr_min,1+dp[j-1])
                    print(f"i={i},j={j}, old_min={old_min}, curr_min={curr_min}")
            dp[i-1] = curr_min
        print(f"dp={dp}")
        return dp[len(s)-1]


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