<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:
        hashmap = defaultdict(set)
        def pals(l, r):
            while l >= 0 and r < len(s) and s[l] == s[r]:
                hashmap[l].add(r)
                l -= 1
                r += 1
        
        for i in range(len(s)):
            pals(i, i)
            pals(i, i+1)

        @lru_cache(None)
        def dfs(ind):
            if ind >= len(s):
                return 0
            
            res = len(s)
            for i in range(ind, len(s)):
                if i in hashmap[ind]:
                    res = min(res, dfs(i+1) + 1)

            return res
        return dfs(0) - 1
210776