<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:
        lens = len(s)
        palin = [[] for i in range(lens)]

        # expand from middle

        for i in range(lens):
            j = i
            k = j
            while j >= 0 and k < lens and s[j] == s[k]:
                palin[j].append(k+1)
                j-=1
                k+=1
            j = i
            k = j+1
            while j >= 0 and k < lens and s[j] == s[k]:
                palin[j].append(k+1)
                j-=1
                k+=1 

        dp = [0] * (lens+1)

        for i in range(lens-1, -1, -1):
            dp[i] = min(dp[j] for j in palin[i])
            dp[i] += 1

        print(palin)
        print(dp)

        return dp[0] - 1
122068