<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=0;
        int j=s.size()-1;
        while(i<j){
            if(s[i]!=s[j]) return false;
            i++;
            j--;
        }
        return true;
    }
    // int f(int i,string s,vector<int>& dp){
    //     if(i==s.size()) return 0;
    //     if(dp[i]!=-1) return dp[i];
    //     string temp="";
    //     int mini=INT_MAX;
    //     for(int j=i;j<s.size();j++){
    //         temp+=s[j];
    //         if(isPalindrome(temp)){
    //             int cost=1+f(j+1,s,dp);
    //             mini=min(mini,cost);
    //         }
    //     }
    //     return dp[i]=mini;
    // }
    int minCut(string s) {
        vector<int>dp(s.size()+1,0);
        //return f(0,s,dp)-1;
        //if(isPalindrome(s)) return 0;
        for(int i=s.size()-1;i>=0;i--){
            string temp="";
            int mini=INT_MAX;
            for(int j=i;j<s.size();j++){
                temp+=s[j];
                if(isPalindrome(temp)){
                    int cost=1+dp[j+1];
                    mini=min(mini,cost);
                }
            }
            dp[i]=mini;
        }
        return dp[0]-1;

    }
};
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