<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();
        int dp[n][n];
        for(int i=0;i<n;i++) dp[i][i]=1;
        for(int i=1;i<n;i++){
            dp[0][i]=1+dp[0][i-1];
           for(int j=i-1;j>0;j--){
               if(s[j]==s[i]){
                if(i==j+1) dp[j][i]=1;
                else if(dp[j+1][i-1]==1) dp[j][i]=1;
                else dp[j][i]=i-j+1;
               }
               else dp[j][i]=i-j+1;
               dp[0][i]=min(dp[0][i],dp[0][j-1]+dp[j][i]);
            //    cout<<j<<' '<<i<<' '<<dp[j][i]<<'\n';
           }
             if(s[0]==s[i]){
                if(i==1) dp[0][i]=1;
                else if(dp[1][i-1]==1) dp[0][i]=1;
            }
        }
        return dp[0][n-1]-1;
    }
};
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