<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 isPal(string &s) {
        int i=0, j= s.size()-1;
        while(i<=j && s[i]==s[j]) {
            i++;
            j--;
        }        
        return (i>j);
    }


    int minCut(string s) {
        int n = s.size();
        vector<int> dp(n,n);

        //dp[i] -> min cuts for substring [i:] 
        string curr = "";
        for(int i=n-1;i>=0;i--) {
            curr += s[i];
            if(isPal(curr)) {dp[i]=0;continue;}
            string temp = "";
            for(int k=i;k<n;k++) {
                temp+=s[k];
                if(!isPal(temp)) continue;
                dp[i] = min(dp[i],1 + dp[k+1]);
            }
        }

        // for(int i=0;i<n;i++) cout << dp[i] << " ";

        return dp[0];
    }
};
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