<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:
        if len(s) <= 1:
            return 0
        L = len(s)
        dp = [[False for _ in range(L)] for i in range(L)]
        
        for r in range(L - 1, -1, -1):
            for c in range(r, L):
                if s[r] == s[c]: 
                    dp[r][c] = c - r <= 2 or dp[r + 1][c - 1]

        if dp[0][-1]:
            print("yes")
            return 0  
        @cache    
        def dfs(s, e):
            if dp[s][e]:
                return 0 
            res = sys.maxsize
            for i in range(s, e):
                if dp[s][i]:
                    res = min(res,  dfs(i + 1, e) + 1)
            return res
        return dfs(0, L -1)
101596