The “Remove Outermost Parentheses” is a popular problem on the Leetcode platform, which tests your string manipulation and stack handling capabilities. In this detailed article, we’ll delve into the problem’s statement, its importance,and approaches to solve it in Python

## Overview:

- Problem Statement
- Importance of the Problem
- Approaches to Solve
- Implementing the Solution in Python
- Testing the Solution
- Conclusion

## 1. Problem Statement:

Given a valid parentheses string `s`

, remove the outermost parentheses of every primitive string in the parentheses string and return `s`

. A parentheses string is primitive if it is non-empty and there does not exist a way to split it into `s = A+B`

, with `A`

and `B`

non-empty parentheses strings.

### Example:

```
Input: s = "(()())(())"
Output: "()()()"
```

## 2. Importance of the Problem:

This problem is a basic test of understanding string manipulations in conjunction with elementary data structures like stacks. It’s a typical representation of problems where you need to deal with nested structures.

## 3. Approaches to Solve:

There are multiple ways to tackle this problem, but two common strategies are:

**Using Stack**: The stack is a natural choice when dealing with problems related to parentheses. We can push each character onto the stack and pop them when certain conditions are met to identify outer parentheses.**Without Stack (Counter Method)**: By using a counter, we can increment it when we see an opening parenthesis and decrement it for a closing one. This way, we can detect outer parentheses without using a stack.

## 4. Implementing the Solution in Python:

### 1. Using Stack:

```
def removeOuterParentheses(s: str) -> str:
stack = []
result = []
primitive_str = []
for char in s:
if char == '(':
stack.append(char)
else:
stack.pop()
primitive_str.append(char)
# When stack is empty, we've found a primitive string.
if not stack:
# Remove the outermost parentheses
result.extend(primitive_str[1:-1])
primitive_str = []
return ''.join(result)
```

### 2. Counter Method:

```
def removeOuterParentheses(s: str) -> str:
count = 0
result = []
for char in s:
# If adding a character to result, avoid outer parentheses
if char == '(' and count > 0:
result.append(char)
if char == ')' and count > 1:
result.append(char)
count += 1 if char == '(' else -1
return ''.join(result)
```

## 5. Testing the Solution:

After implementing the solution, it’s crucial to test it to ensure correctness:

```
test_string = "(()())(())"
print(removeOuterParentheses(test_string)) # Expected output: "()()()"
test_string2 = "(()())(())(()(()))"
print(removeOuterParentheses(test_string2)) # Expected output: "()()()()(())"
test_string3 = "()()"
print(removeOuterParentheses(test_string3)) # Expected output: ""
```

## 6. Conclusion:

The “Remove Outermost Parentheses” problem on Leetcode is an excellent way to familiarize oneself with string manipulations and the stack data structure. While the problem might appear straightforward, careful consideration ensures that the solution is not only correct but also efficient. Python offers intuitive ways to approach this problem, making it accessible to beginners and experienced coders alike.