<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>

#define ll long long
class Solution {
public:
    int minCut(string s) {
        ll n=s.size();
        ll pal[n][n],dp[n+1];
        memset(pal,0,sizeof(pal));
        memset(dp,0x3f,sizeof(dp));

        for(ll i=0;i<n;i++)
            for(ll j=0;j<n;j++)
                pal[i][j]=(j<=i);

        for(ll len=1;len<=n;len++){
            for(ll l=0;l+len<n;l++){
                ll r=l+len;
                pal[l][r]=pal[l+1][r-1]&&(s[l]==s[r]);
            }
        }
        dp[n]=0;
        for(ll i=n-1;i>=0;i--){
            for(ll j=i;j<n;j++){
                if(pal[i][j])
                    dp[i]=min(dp[i],dp[j+1]+1);
            }
        }
        return dp[0]-1;
    }
};
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