1021. Remove Outermost Parentheses_S.toCharArray()

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 + ... + Pk, where Pi are primitive valid parentheses strings.

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

Example One:

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



Example Two:

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



Example Three:


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

Solution:

class Solution {
    public String removeOuterParentheses(String S) {
        StringBuilder  s = new StringBuilder();
        int open = 0;
        for(char k : S.toCharArray()){
            if(k == '('  && open++ > 0){
                s.append(k);
            }
            if(k == ')'  && open-- > 1){
                s.append(k);
            }
        }
        return s.toString();
    }
}

Hint:

Logics:

• Have a StringBuilder that will store the result
• only need to find the () 
• then return s.toString() to output the result

Tips:

• notice the differnce between open++ and ++open
• understand the logic
• use s.toCharString() to get each char
• need to  return s.toString() to output the result(because char can't be converted to Stirng directly)