<p>Given a balanced parentheses string <code>s</code>, return <em>the <strong>score</strong> of the string</em>.</p>

<p>The <strong>score</strong> of a balanced parentheses string is based on the following rule:</p>

<ul>
	<li><code>&quot;()&quot;</code> has score <code>1</code>.</li>
	<li><code>AB</code> has score <code>A + B</code>, where <code>A</code> and <code>B</code> are balanced parentheses strings.</li>
	<li><code>(A)</code> has score <code>2 * A</code>, where <code>A</code> is a balanced parentheses string.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;()&quot;
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;(())&quot;
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;()()&quot;
<strong>Output:</strong> 2
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= s.length &lt;= 50</code></li>
	<li><code>s</code> consists of only <code>&#39;(&#39;</code> and <code>&#39;)&#39;</code>.</li>
	<li><code>s</code> is a balanced parentheses string.</li>
</ul>

class Solution {
public:
    int recur(int i,int j,vector<vector<int>> &dp,string &s){
        if(i>j) return 0;
        if((j-i)==1) return dp[i][j]=1;
        if(dp[i][j]!=-1) return dp[i][j];
        int c=0;
        for(int k=i;k<=j;k++){
            if(s[k]=='(') c++;
            else c--;
            if(c==0){
                int l1=max(1,2*recur(i+1,k-1,dp,s));
                int l2=recur(k+1,j,dp,s);
                dp[i][j]=l1+l2;
                break;
            }
        }
        return dp[i][j];
    }
    int scoreOfParentheses(string s) {
        int n=s.size();
        vector<vector<int>> dp(n,vector<int>(n,-1));
        return recur(0,n-1,dp,s);
    }
};
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