<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)
        if N == 1:
            return 0
        dp = dict()
        def helper(i, j):
            while i >= 0 and j < N and s[i] == s[j]:
                if i in dp:
                    dp[i].add(j)
                else:
                    dp[i] = set([j])
                i -= 1
                j += 1
        
        for k in range(N):
            helper(k, k)
            helper(k, k+1)
        cache = {}
        def solve(i):
            if i in cache:
                return cache[i]
            if i > N-1:
                return -1
            if i == N-1:
                return 0
            sol = N - i -1
            if N-1 in dp[i]:
                return 0
            for j in dp[i]:
                sol = min(1 + solve(j+1), sol)
            cache[i] = sol
            return sol

        return solve(0)
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