1021. Remove Outermost Parentheses

Description

A valid parentheses string is either empty "", "(" + A + ")", or A + B, where A and B are valid parentheses strings, and + represents string concatenation.

For example, "", "()", "(())()", and "(()(()))" are all valid parentheses strings.

A valid parentheses string s is primitive if it is nonempty, and there does not exist a way to split it into s = A + B, with A and B nonempty valid parentheses strings.

Given a valid parentheses string s, consider its primitive decomposition: s = P₁ + P₂ + ... + P_k, where P_i are primitive valid parentheses strings.

Return s after removing the outermost parentheses of every primitive string in the primitive decomposition of s.

Example 1:

Input: s = "(()())(())"
Output: "()()()"
Explanation: 
The input string is "(()())(())", with primitive decomposition "(()())" + "(())".
After removing outer parentheses of each part, this is "()()" + "()" = "()()()".

Example 2:

Input: s = "(()())(())(()(()))"
Output: "()()()()(())"
Explanation: 
The input string is "(()())(())(()(()))", with primitive decomposition "(()())" + "(())" + "(()(()))".
After removing outer parentheses of each part, this is "()()" + "()" + "()(())" = "()()()()(())".

Example 3:

Input: s = "()()"
Output: ""
Explanation: 
The input string is "()()", with primitive decomposition "()" + "()".
After removing outer parentheses of each part, this is "" + "" = "".

Constraints:

1 <= s.length <= 10⁵
s[i] is either '(' or ')'.
s is a valid parentheses string.

Solution

remove-outermost-parentheses.py

class Solution:
    def removeOuterParentheses(self, s: str) -> str:
        opened = 0
        res = []

        for x in s:
            if opened > 0:
                res.append(x)

            if x == "(":
                opened += 1
            else:
                opened -= 1
                if opened == 0:
                    res.pop()

        return "".join(res)