<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:
        dp = [[False for i in range(len(s))] for i in range(len(s))]
        for l in range(len(s)):
            for start in range(len(s)-l):
                i = start
                j = start+l
                if(l==0):
                    dp[i][j] = True
                elif(l==1):
                    dp[i][j] = (s[i] == s[j])
                else:
                    dp[i][j] = (s[i] == s[j]) and dp[i+1][j-1]
        valid_inds = []
        for j in range(len(s)):
            curr_inds = []
            for i in range(j+1):
                if(dp[i][j]):
                    curr_inds.append(i)
            valid_inds.append(curr_inds)
        
        sols = [float("inf") for i in range(len(s))]
        sols[0] = 0
        for i in range(len(sols)):
            for ind in valid_inds[i]:
                if(ind == 0):
                    sols[i] = 0
                else:
                    sols[i]= min(sols[ind-1] + 1, sols[i])
        return sols[-1]
        
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