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Java Solution for LeetCode algorithm problems, continually updating
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1021\. Remove Outermost Parentheses
Easy
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 <= 105
* `s[i]` is either `'('` or `')'`.
* `s` is a valid parentheses string.