<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 isPalindrome(string &s,int i,int j){
        int st=i;
        int e=j;
        while(st<e){
            if(s[st]!=s[e])return false;
            st++;
            e--;
        }
        return true;
    }
    int f(int i,int j,string &s, vector<vector<int>>&dp){
        if(i>=j || isPalindrome(s,i,j))return 0;
       if(dp[i][j]!=-1)return dp[i][j];
        string ans="";
        int z=1e9;
        for(int ind=i;ind<j;ind++){
            ans+=s[ind];
            if(isPalindrome(s,i,ind)){
                z=min(z,1+f(ind+1,j,s,dp));
            }

        }
        ans+=s[j];
   
        return dp[i][j]=z;
    }
    int minCut(string s) {
      
        int n=s.size();
        if(isPalindrome(s,0,n-1))return 0;
        vector<vector<int>>dp(n,vector<int>(n,-1));
        return f(0,n-1,s,dp);
    }
};
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