LeetCode: Generate Parentheses

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

Intuition: The simplest way to tackle this would be to generate all 2^(2n) strings containing ‘(’ and ‘)’, then validate each of this strings if they are valid or not. However this is not an optimal solution since time complexity would be O((2^(2n)) * n) and space complexity would also be O((2^(2n)) * n).

Optimal Solution(Backtracking): In this approach, we add ‘(’ and ‘)’ only when we know the sequence getting created will be valid. The code for the same is below:

class Solution {
    public List<String> generateParenthesis(int n) {
        List<String> res = new ArrayList();
        
        backtrack(res, n, new StringBuilder(), 0, 0);
        return res;
    }
    
    public void backtrack(List<String> res, int n, StringBuilder curr, int open, int close) {
        if(curr.length() == 2*n){
            //valid string added to result
            res.add(curr.toString());
            return;
        } 
        
        if(open < n) {
            //since open paranthesis are still left
            curr.append("(");
            backtrack(res, n, curr, open+1, close);
            curr.deleteCharAt(curr.length()-1);
        }
        
        if(close < open){
            //since closing paranthesis are less
            curr.append(")");
            backtrack(res, n, curr, open, close+1);
            curr.deleteCharAt(curr.length()-1);
        }
    }
}