LeetCode 1021 - Remove Outermost Parentheses

https://leetcode.com/problems/remove-outermost-parentheses/

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 = P_1 + P_2 + ... + P_k, where P_iare primitive valid parentheses strings.

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

Example 1:

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

Example 2:

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

Example 3:

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

Note:

1. S.length <= 10000
2. S[i] is "(" or ")"
3. S is a valid parentheses string

https://leetcode.com/problems/remove-outermost-parentheses/discuss/270022/JavaC%2B%2BPython-Count-Opened-Parenthesis

opened count the number of opened parenthesis.
Add every char to the result,
unless the first left parenthesis,
and the last right parenthesis.

Time Complexity:

O(N) Time, O(N) space

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

X. https://leetcode.com/problems/remove-outermost-parentheses/discuss/270566/My-Java-3ms-Straight-Forward-Solution-or-Beats-100

    public String removeOuterParentheses(String S) {
        
        StringBuilder sb = new StringBuilder();
        int open=0, close=0, start=0;
        for(int i=0; i<S.length(); i++) {
            if(S.charAt(i) == '(') {
                open++;
            } else if(S.charAt(i) == ')') {
                close++;
            }
            if(open==close) {
                sb.append(S.substring(start+1, i));
                start=i+1;
            }
        }
        return sb.toString();
    }