Here, We see Generate Parentheses LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Generate Parentheses LeetCode Solution

Problem Statement

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

Example 1:
Input: n = 3
Output: [“((()))”,”(()())”,”(())()”,”()(())”,”()()()”]

Example 2:
Input: n = 1
Output: [“()”]

Generate Parentheses Leetcode Solution C++

class Solution {
public:
    vector<string> generateParenthesis(int n) {
        int open = n;
        int close = n;
        vector<string> ans;
        string op = "";
        solve(op, open, close, ans);
        return ans;        
    }
        void solve(string op, int open, int close, vector<string> &ans){
        if(open == 0 && close == 0){
            ans.push_back(op);
            return;
        } 
        if(open == close){
            string op1 = op;
            op1.push_back('(');
            solve(op1, open-1, close, ans);
        }
        else if(open == 0){
            string op1 = op;
            op1.push_back(')');
            solve(op1, open, close-1, ans);
        }
        else if(close == 0){
            string op1 = op;
            op1.push_back('(');
            solve(op1, open-1, close, ans);
        }
        else{
            string op1 = op;
            string op2 = op;
            op1.push_back('(');
            op2.push_back(')');
            solve(op1, open-1, close, ans);
            solve(op2, open, close-1, ans);
        }
    }
};Code language: JavaScript (javascript)

Generate Parentheses Leetcode Solution Java

class Solution {
    public List<String> generateParenthesis(int n) {
        List<String> res = new ArrayList<String>();
        recurse(res, 0, 0, "", n);
        return res;
    }
    public void recurse(List<String> res, int left, int right, String s, int n) {
        if (s.length() == n * 2) {
            res.add(s);
            return;
        }
        if (left < n) {
            recurse(res, left + 1, right, s + "(", n);
        }
        if (right < left) {
            recurse(res, left, right + 1, s + ")", n);
        }
    }
}Code language: JavaScript (javascript)

Generate Parentheses Solution JavaScript

var generateParenthesis = function(n) {
  const output = [];
  const dfs = (str, open, close) => {
    if (open > close) {
      return;
    }
    if (open === 0 && close === 0) {
      output.push(str);
      return;
    }
    if (open > 0) {
      dfs(`${str}(`, open - 1, close);
    }
    if (close > 0) {
      dfs(`${str})`, open, close - 1);
    }
  };
  dfs('', n, n);
  return output;
};Code language: JavaScript (javascript)

Generate Parentheses Solution Python

class Solution(object):
    def generateParenthesis(self, n):
	    def dfs(left, right, s):
		    if len(s) == n * 2:
			    res.append(s)
			    return 
		    if left < n:
			    dfs(left + 1, right, s + '(')
		    if right < left:
			    dfs(left, right + 1, s + ')')
	    res = []
	    dfs(0, 0, '')
	    return res