# 1021. 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 "" + "" = "".
```

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.

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