Problem #22: Generate Parentheses

Intuition & Strategy Breakdown

Use backtracking to construct valid combinations. Add an open bracket '(' if open count < n, and add a close bracket ')' if close count < open count.

Python Solution

def generateParenthesis(n):
    res = []
    def backtrack(S, left, right):
        if len(S) == 2 * n:
            res.append(S)
            return
        if left < n:
            backtrack(S + "(", left + 1, right)
        if right < left:
            backtrack(S + ")", left, right + 1)
    backtrack("", 0, 0)
    return res

Java Solution

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