<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:
    bool isValid(string &s)
    {
        stack<char> parantheses;

        for (char &c : s)
        {
            if (c == '(') parantheses.push(c);
            else
            {
                if (parantheses.empty()) return false;
                if (parantheses.top() != '(') return false;
                else parantheses.pop();
            }
        }

        return (parantheses.empty());
    }
    vector<string> generateParenthesis(int n) 
    {  
        vector<string> combinations;
        unordered_set<string> seen;

        string s = "";

        for (int i = 0; i < n; i++)
        {
            s.push_back('(');
        }
        for (int i = 0; i < n; i++)
        {
            s.push_back(')');
        }

        combinations.push_back(s);
        seen.insert(s);


        while (next_permutation(s.begin(), s.end()))
        {
            if (seen.find(s) == seen.end()) 
            {
                if (isValid(s)) combinations.push_back(s);
                seen.insert(s);
            }
            else break;
        }

        return combinations;
    }
};
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