Remove Outermost Parentheses || String

[Question]: 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.

Return s after removing the outermost parentheses of every primitive string in the primitive decomposition of s.

Example 1:

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

Example 2:

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

Example 3:

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

// TC: O(n)
// SC: O(_)
func removeOuterParentheses(_ s: String) -> String {
    var outputStr = ""
    var opened = 0
    for c in s {
        if c == "(" {
            if opened > 0 {
                outputStr.append(c)
            }
            opened+=1
        }
        if c == ")" {
            if opened > 1  {
                outputStr.append(c)
            }
            opened-=1
        }
    }
    return outputStr
}

let parnthString = "(()())(())"
let opRemovedLastParath = removeOuterParentheses(parnthString)
print(" removed last Paranth", opRemovedLastParath) // ()()()

Dry Run: for (  ( )  ( )  )  (  (  )  )

Input	Opened Current	Opened After	Output
(	0	1	“”
(	1	2	(
)	2	1	(  )
(	1	2	( ) (
)	2	1	( ) ( )
)	1	0	( ) ( )
(	0	1	( ) ( )
(	1	2	( ) ( ) (
)	2	1	( ) ( )  ( )
)	1	0	( ) ( )  ( )