Remove One Outer Pair from Every Primitive Parentheses Component
Company: Amazon
Role: Software Engineer
Category: Coding & Algorithms
Difficulty: easy
Interview Round: Other
### Problem
A valid parentheses string is empty, a matching pair around a valid string, or a concatenation of valid strings. A nonempty valid string is primitive when it cannot split into two nonempty valid strings. Given valid `s`, decompose it into primitive components and remove each component's outermost pair. Return their remaining interiors concatenated in order.
```cpp
class Solution {
public:
string removeOuterParentheses(string s);
};
```
### Examples
**Example 1:**
```text
Input: s = "((()))()"
Output: "(())"
Explanation: The primitive components are((())) and(); their interiors are(()) and the empty string.
```
**Example 2:**
```text
Input: s = "(())(())"
Output: "()()"
Explanation: Each primitive component loses its outer pair and contributes().
```
### Constraints
- `1 <= s.length <= 10^5`.
- Every character is `(` or `)`.
- s is a valid parentheses string.
### Contract considerations
The input may consist of many primitive components, and each loses its own outer pair.
```hint Read the boundary condition
Depth returning to zero identifies the boundary between primitive components.
```
### Boundary reminders
The input may consist of many primitive components, and each loses its own outer pair.
Overview: Remove One Outer Pair from Every Primitive Parentheses Component. A valid parentheses string is empty, a matching pair around a valid string, or a concatenation of valid strings.
