<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.length();
        vector<int>dp(n,0);
        vector<vector<int>>is_pal(n, vector<int>(n,0));
        for(int i=n-1 ; i>=0 ; i--){
            for(int j=i ; j<n ; j++){
                if(i==j) is_pal[i][j]=1;
                else{
                    if(s[i]==s[j] && (j-i==1 || is_pal[i+1][j-1])){
                        is_pal[i][j]=1;
                    }
                }
            }
        }
        for(int i=n-2 ; i>=0 ; i--){
            if(!is_pal[i][n-1]){
                int e=INT_MAX;
                for(int j=i ; j<=n-2 ; j++){
                    // if(i==0 && j==1) cout<<1+dp[j+1]<<endl;
                    if(is_pal[i][j]) e=min(e, 1+dp[j+1]);
                }
                // if(i==0) cout<<"ha "<<e<<endl;
                dp[i]=e;
            }
        }
        // cout<<"haaaaa"<<is_pal[0][1]<<endl;
        return dp[0];

    }
};
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