<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 palindrome(string &s)
    {
        int n=s.length();
        // bool flag=true;
        for(int i=0; i<n/2; i++)
        {
            if(s[i]!=s[n-i-1])
            return false;
        }
        return true;
    }
    int f(int i, string &s, vector<int> &dp)
    {
        int n=s.length();
        if(i>=n)
        return 0;
        if(dp[i]!=-1)
        return dp[i];
        int mini=1e9;
        string t="";
        int cnt=0;
        for(int j=i; j<n; j++)
        {
            // if(cnt==0 && palindrome(t))
            // {
            //     return 0;
            // }
            t+=s[j];
            if(palindrome(t))
            {
                mini=min(mini,1+f(j+1,s,dp));
            }
            // t+=s[i];
        }
        return dp[i]=mini;
    }
    int minCut(string s) {
        int n=s.length();
        vector<int> dp(n,-1);
        return f(0,s,dp)-1;
    }
};
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