Remove Outermost Parentheses - LeetCode solutions

Problem

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 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 <= 10^5
s[i] is either '(' or ')'.
s is a valid parentheses string.

Solution (Java)

class Solution {
    public String removeOuterParentheses(String s) {
        List<String> primitives = new ArrayList<>();
        int i = 1;
        while (i < s.length()) {
            int initialI = i - 1;
            int left = 1;
            while (i < s.length() && left > 0) {
                if (s.charAt(i) == '(') {
                    left++;
                } else {
                    left--;
                }
                i++;
            }
            primitives.add(s.substring(initialI, i));
            i++;
        }
        StringBuilder sb = new StringBuilder();
        for (String primitive : primitives) {
            sb.append(primitive, 1, primitive.length() - 1);
        }
        return sb.toString();
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).