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 "()()" + "()" = "()()()".

"(()())(())(()(()))"

"()()()()(())"

The input string is "(()())(())(()(()))", with primitive decomposition "(()())" + "(())" + "(()(()))".
After removing outer parentheses of each part, this is "()()" + "()" + "()(())" = "()()()()(())".

"()()"

""

The input string is "()()", with primitive decomposition "()" + "()".
After removing outer parentheses of each part, this is "" + "" = "".

Note:

1. S.length <= 10000
2. S[i] is "(" or ")"
3. S is a valid parentheses string

class Solution {
    public String removeOuterParentheses(String S) {
        int lv = 0;
        String result = "";
        
        for(char c : S.toCharArray()){
            if(c == '('){
                if(lv > 0){
                    result += c;
                }
                lv ++;
                
            } else if(c == ')'){
                lv--;
                if(lv > 0){
                    result += c;
                }
            }
        }
        return result;
    }
}

* Using a StringBuilder will somehow improve the performance.

Thoughts:

Since we can assume the input is valid, just use a counter to mark the depth. Send char to result only when depth >= 1

If I’m a Interviewer

1. What if the input string is not always valid?
   • ref: LC20
2. What if we upgrade parentheses to brackets ()[]{} ?
   • Use Stack and Stack.size() instead of level counter
3. What if we remove inner most (max depth) parentheses instead?
   • Go through twice, 1) find max depth, 2) copy to result and skip max depth.

Well we can always construct an array of ints (levels) as a form of representation.

For example, “(()())(())(()(()))” -> [1,2,2,1,2,1,2,2,3]

Then we can easily remove any levels (outer/inner/somewhere in the middle).

• Remove Outer: remove 1, everything else —
• Remove Inner: remove max
• Remove Level2: remove 2, everything > 2 decrease by 1.