<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 {
public:
    int minCut(string s) {
        int n = s.size();
        vector<int>dp(n + 1, INT_MAX);
        dp[0]=0;
        vector<bool>prevp(n, 0), curp(n, 0);
        prevp[0] = 1;
        curp[0] = 1;
        for(int i=1;i<=n;i++){
            for(int j=0;j<i;j++){
                if(j==i-1||prevp[i-j-2] && s[i-1] == s[j]){
                    curp[i-j]=1;
                    dp[i] = min(dp[i], dp[j] + 1);
                }
            }
            prevp = curp;
            curp.assign(n,0);
            curp[0] = 1;
        }
        return dp[n] - 1;
    }
};
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