[LeetCode] 22. Generate Parentheses

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
For example, given n = 3, a solution set is:
[
  "((()))",
  "(()())",
  "(())()",
  "()(())",
  "()()()"
]

Thought process:
The idea is to recursively add and remove parentheses in a string. When the string's length reaches 2 * n, we have our base case and we can add the string to result list.
To make sure the parentheses are well-formed, we can use two counters: open and close, which tracks open parenthesis and close parenthesis respectively. When open < n, we can add more open parentheses; when close < open, we can add more close parentheses.

Solution:
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class Solution {
public:
    vector<string> generateParenthesis(int n) {
        vector<string> parentheses;
        generateParentheses(n, "", parentheses, 0, 0);
        return parentheses;
    }
    
    void generateParentheses(int n, string s, vector<string>& parentheses, int open, int close) {
        if (s.length() == 2 * n) {
            parentheses.push_back(s);
            return;
        }
        
        if (open < n) {
            generateParentheses(n, s + "(", parentheses, open + 1, close);
        }
        
        if (close < open) {
            generateParentheses(n, s + ")", parentheses, open, close + 1);
        }
    }
};

It's okay to overload a function with a different return type.

Time complexity:

Same as combinations: O(n!).
Location: Philadelphia, PA, USA

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