<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:
    bool checkPalindrme(string &s,int l,int m){
        int i=l,j=m;
        while(i<=j){
            if(s[i]!=s[j]){
                return false;
            }
            i+=1;
            j-=1;
        }
        return true;
    }
    int rec(int i,int j,string &s,vector<vector<int>> &dp){
        if(i>=j-1) return 0;
        if(dp[i][j]!=-1) return dp[i][j];
        int minCuts=INT_MAX;
        for(int k=i;k<j;k++){
            int right=0;
            if(checkPalindrme(s,i,k)){
               right=rec(k+1,j,s,dp);
               if(k==j-1) right-=1;
               minCuts=min(minCuts,1+right);
            } 
        
        }
        return dp[i][j]=minCuts;
    }
    int minCut(string s) {
        vector<vector<int>> dp(s.size()+1, vector<int>(s.size()+1,-1));
        return rec(0,s.size(),s,dp);
    }
};
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