<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:
        dp = [0]*len(s)
        start_index = {i: set() for i in range(len(s))}
        for i in range(len(s)):
            c = 0
            while(i-c >= 0 and i+c < len(s) and s[i-c] == s[i+c]):
                start_index[i+c].add(i-c)
                c+=1
            c = 0
            while(i-c-1 >= 0 and i+c < len(s) and s[i-c-1] == s[i+c]):
                start_index[i+c].add(i-c-1)
                c+=1
        for i in range(1, len(s)):
            m = dp[i-1]+1
            for s_index in start_index[i]:
                m = min(m, 0 if s_index == 0 else dp[s_index-1]+1)
            dp[i] = m
        return dp[-1]


        
176658