<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:
        n = len(s)
        t = [[-1 for i in range(n)] for _ in range(n)]
        dp = [[None for i in range(n)] for _ in range(n)]
        def isPalindrome(string, i, j):
            if i >= j:
                return True
            if dp[i][j] != None:
                return dp[i][j]
            dp[i][j] = string[i] == string[j] and isPalindrome(string, i + 1, j - 1)
            return dp[i][j]
        def solve(i, j):
            if i >= j:
                return 0
            if isPalindrome(s, i, j):
                return 0
            if t[i][j] != -1:
                return t[i][j]
            t[i][j] = float('inf')
            for k in range(i, j):
                if isPalindrome(s, i, k):
                    t[i][j] = min(t[i][j], 1 + solve(i, k) + solve(k + 1, j))
            return t[i][j]
        return solve(0, n - 1)
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