<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
        isPalin = lambda x: x == x[::-1]
        cache = dict()
        # calculate and cache all substring is palindrome or not
        for r in range(len(s) + 1):
            for l in range(r):
                cache[(l, r)] = isPalin(s[l:r])
        # save down the minimum number of cut for string[0:r] in dp[r]
        dp = [float('inf') for i in range(len(s) + 1)]
        dp[0] = 0
        for r in range(1, len(s) + 1):
            if cache[(0, r)]:
                # if string[0:r] is palindrome, set dp[r] = 0 and skip loop
                dp[r] = 0
                continue
            for l in range(1, r):
                # if string[l:r] is palindrome, set dp[r] = min(dp[r], dp[l] + 1)
                # dp[l] is the best number of cut in string[0:l]
                # number 1 is come from string[l:r] is palindrome
                if cache[(l, r)]:
                    dp[r] = min(dp[r], dp[l] + 1)
        return dp[-1]
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