# Leetcode - Remove Outermost Parentheses Solution

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_i`

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: "(()())(())"
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:**

`S.length <= 10000`

`S[i]`

is`"("`

or`")"`

`S`

is a valid parentheses string

## Solution in Python

```
class Solution:
def removeOuterParentheses(self, S: str) -> str:
left = 0
right = 0
curr_stack = []
main_stack = []
for i in S:
if i == "(":
left+=1
curr_stack.append("(")
else:
right+=1
curr_stack.append(")")
if left==right:
main_stack.extend(curr_stack[1:-1])
curr_stack = []
return "".join(main_stack)
```