# LeetCode – Stack – Remove Outermost Parentheses

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 "" + "" = "".```
``public String removeOuterParentheses(String S) {        String result = "";        Stack<String> s1 = new Stack<String>();        Stack<String> s2 = new Stack<String>();                for(int i=0;i<S.length();i++){            //System.out.println(S.charAt(i));            if(s1.size() == 0){                s1.push(S.charAt(i)+"");            }            else if(s1.size() > 0){                if(("(").equals(S.charAt(i)+"")){                    s2.push(S.charAt(i)+"");                    result = result+S.charAt(i);                }                else if(s2.size() > 0 && (")").equals(S.charAt(i)+"")){                    s2.pop();                    result = result+S.charAt(i);                }                else{                    s1.pop();                }            }        }        return result;    }``