<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 check(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 n;
    int recurssion(string &s,int ind,vector<int>&dp){
        if(ind==n){
            return 0;
        }
        if(dp[ind]!=-1) return dp[ind];
        int mini=INT_MAX;
        string temp="";
        for(int i=ind;i<n;i++){
            temp.push_back(s[i]);
            if(check(temp)){
                int cost=1+recurssion(s,i+1,dp);
                mini=min(mini,cost);
            }
        }
        return dp[ind]=mini;
    }
    int minCut(string s) {
        n=s.size();
        vector<int>dp(n,-1);
        return recurssion(s,0,dp)-1;
    }
};
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