<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 generate(vector<string> &ans, int num, int open, int close, string res, char ch)
    {
        if(open == num && close == num)
        {
            for(int i=0; i<ans.size(); i++)
            {
                if (ans[i] == res)
                    return;
            }
            ans.push_back(res);
            return;
        }

        if((open >= close) && (open <= num))
        {
            if((ch == '(')&& (open < num))
            {
                res = res+ch;
                open++;
                generate(ans, num, open, close, res, '(');
                generate(ans, num, open, close, res, ')');
            }
            else if((ch == ')')&& (close < open))
            {
                res = res+ch;
                close++;
                generate(ans, num, open, close, res, '(');
                generate(ans, num, open, close, res, ')');
            }
        }
        return;
    }
  
    vector<string> generateParenthesis(int n) {
        
        vector<string> ans;
        generate(ans,n, 0, 0, "", '(');
        return ans;
    }
};
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