<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:
        def isPalindrome(s):
            n = len(s)
            dp = [[False] * n for _ in range(n)]
            for i in range(n-1, -1, -1):
                for j in range(i, n):
                    if i == j:
                        dp[i][j] = True
                    elif j - i == 1:
                        dp[i][j] = (s[i] == s[j])
                    else:
                        dp[i][j] = (s[i] == s[j] and dp[i+1][j-1])
            return dp

        dp = isPalindrome(s)
        n = len(s)
        res = float('inf')

        @cache
        def minPartition(start, res):
            if dp[start][n-1]:
                return 0

            for i in range(start + 1, n+1):
                if dp[start][i-1]:
                    res = min(res, 1 + minPartition(i, res))
            return res
        res = minPartition(0, res)
        return res
122068