<p>Given <code>n</code> pairs of parentheses, write a function to <em>generate all combinations of well-formed parentheses</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> n = 3
<strong>Output:</strong> ["((()))","(()())","(())()","()(())","()()()"]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> n = 1
<strong>Output:</strong> ["()"]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 8</code></li>
</ul>

class Solution {
public:
    void find(int ind,int n , vector<string> &ans, string s,int &cnt1,int &cnt2){
        if(ind==n){
            ans.push_back(s);
            return;
        }       
        if(cnt1==n/2){
            s[ind] = ')';
            cnt2++;
            find(ind+1,n,ans,s,cnt1,cnt2);
            cnt2--;
        }
        else{
            s[ind] = '(';
            cnt1++;
            find(ind+1,n,ans,s,cnt1,cnt2);
            cnt1--;
            if(cnt1==cnt2) return;
            s[ind] = ')';
            cnt2++;
            
            find(ind+1,n,ans,s,cnt1,cnt2);
            cnt2--;
        }


    }
    vector<string> generateParenthesis(int n) {
        vector<string> ans;
        string s(2*n,'.');
        s[0] ='(';
        int cnt1=1;
        int cnt2=0;
        find(1,2*n,ans,s,cnt1,cnt2);
        cout<<cnt1<<" "<<cnt2;
        return ans;
        
    }
};
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