Leetcode 22 : Generate Parentheses Solution

Problem Statement could be found @https://leetcode.com/problems/generate-parentheses/

Approach -

1.Take a char array of size 2*n.
2. At every value of n try putting '(' or ')'. Reduce count of open brackets and closed brackets
3. If both open brackets & close brackets count reduce to 0,then it means we have  got one combination
4. A condition to take care is - never count of open brackets is greater than close brackets as it would result in wrong code formation

Code Snippet-

class Solution {
public:
    vector<string> retVec;
   
    void helper(int open ,int close, int n, char *x,int index){
        if ((open == 0) && (close ==0)){
            retVec.push_back(x);
            return;
        }
        //results in wrong formation
        if (open > close){
            return;
        }
       
        if (open > 0){
            x[index] = '(';
            helper((open-1),close, n, x, index+1);
        }
       
        if (close > 0){
            x[index] = ')';
            helper(open,close-1, n, x, index+1);
        }
    }
   
    vector<string> generateParenthesis(int n) {
        char *ptr = new char[2*n+1];
        ptr[2*n] = 0;
        helper(n,n,n,ptr,0);
        return retVec;
    }
};