Generate Parentheses LeetCode Solution

Problem – Generate Parentheses LeetCode Solution

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: ["()"]

Constraints:

• 1 <= n <= 8

Generate Parentheses LeetCode Solution in Java

 public List<String> generateParenthesis(int n) {
        List<String> list = new ArrayList<String>();
        backtrack(list, "", 0, 0, n);
        return list;
    }
    
    public void backtrack(List<String> list, String str, int open, int close, int max){
        
        if(str.length() == max*2){
            list.add(str);
            return;
        }
        
        if(open < max)
            backtrack(list, str+"(", open+1, close, max);
        if(close < open)
            backtrack(list, str+")", open, close+1, max);
    }

Generate Parentheses LeetCode Solution in C++

class Solution {
public:
    vector<string> generateParenthesis(int n) {
        vector<string> res;
        addingpar(res, "", n, 0);
        return res;
    }
    void addingpar(vector<string> &v, string str, int n, int m){
        if(n==0 && m==0) {
            v.push_back(str);
            return;
        }
        if(m > 0){ addingpar(v, str+")", n, m-1); }
        if(n > 0){ addingpar(v, str+"(", n-1, m+1); }
    }
};

Generate Parentheses LeetCode Solution in Python

class Solution(object):
    def generateParenthesis(self, n):
        """
        :type n: int
        :rtype: List[str]
        """
        dp = [[] for i in range(n + 1)]
        dp[0].append('')
        for i in range(n + 1):
            for j in range(i):
                dp[i] += ['(' + x + ')' + y for x in dp[j] for y in dp[i - j - 1]]
        return dp[n]

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