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# 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){
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.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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