<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:
    string getStr(vector<char>& vc) {
        string res = "";
        for(auto ele: vc) {
            res += ele;
        }
        return res;
    }
    
    void backtrack(int openCnt, int closeCnt, int& n, vector<string>& res, vector<char>& vc) {
        if(openCnt == closeCnt && openCnt == n) {
            res.emplace_back(getStr(vc));
            return;
        }
        if(openCnt < n) {
            vc.emplace_back('(');
            backtrack(openCnt + 1, closeCnt, n, res, vc);
            vc.pop_back();
        }
        if(closeCnt < openCnt) {
            vc.emplace_back(')');
            backtrack(openCnt, closeCnt + 1, n, res, vc);
            vc.pop_back();
        }
    }
    
    vector<string> generateParenthesis(int n) {
        vector<string> res;
        vector<char> vc;
        backtrack(0, 0, n, res, vc);
        return res;
    }
};
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