[LeetCode] 22. Generate Parentheses_Medium tag: backtracking

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

 

Idea:

backtracking. 

Code1, generate all pairs with 2 * n, then check whether they are valid parentheses.

T: O(2^(2n) * n)

S: O(2^(2n) * n)

class Solution:
    def generateParentheses(self, n):
        ans = []
        def helper(ans, temp, n):
            if len(temp) == 2 * n and self.isValid(temp):
                ans.append(temp)
            elif len(temp) < 2 * n:
                helper(ans, temp + '(', n)
                helper(ans, temp + ')', n)
        helper(ans, "", n)
    
    def isValid(self, s):
        count = 0
        for c in s:
            if c == '(':
                count += 1
            else:
                count -= 1
            if count < 0:
                return False
        return count == 0

 

Idea 2, we know that with 2 * n length parentheses, we need to have n '(' and n ')', so in the helper function, add left and right to count the number of '(' and ')'. 

Time: O(2^(2n) * n /n/ n ^ (1/2)) = 4^n/n ^ (1/2)

Space : O(2^(2n) * n /n/ n ^ (1/2)) = 4^n/n ^ (1/2)

class Solution:
    def generateParentheses(self, n):
        ans = []
        def helper(ans, temp, left, right, n):
            if len(temp) == 2 * n:
                ans.append(temp)
            elif len(temp) < 2 * n:
                if left < n:
                    helper(ans, temp + '(', left + 1, right, n)
                if right < left:
                    helper(ans, temp + ')', left, right + 1, n)
        helper(ans, "", 0, 0, n)
        return ans