<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 {
    int n;
    string s;
    vector<int> dp;
    vector<vector<bool>> done, pal;
    bool isPal(int l,int r){
        if(l>=r) return true;
        if(done[l][r]) return pal[l][r];
        done[l][r]=true;
        return pal[l][r] = (s[l]==s[r]) ? isPal(l+1,r-1) : false;
    }
    int rec(int index){
        if(index>=n) return 0;
        if(dp[index]!=-1) return dp[index];
        int ans =1e9;
        for(int i=index;i<n;i++){
            if(isPal(index,i)){
                ans = min(ans, 1+rec(i+1));
            }
        }
        return dp[index]=ans;
    }
public:
    int minCut(string S) {
        s=S;
        n= s.size();
        dp.assign(n,-1);
        pal.assign(n,vector<bool>(n,false));
        done.assign(n,vector<bool>(n,false));
        return rec(0)-1;
    }
};
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