remove-outermost-parentheses.js

/*
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 = P1 + P2 + ... + Pk, where Pi 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: 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 "()()" + "()" + "()(())" = "()()()()(())".

*/

var removeOuterParentheses = function (s) {
    let stack = []
    let str = ""

    for (let i = 0; i < s.length; i++) {
        if (s[i] === "(") {
            stack.push(s[i])
            if(stack.length > 1) str += s[i]
        }
        else if (s[i] === ")") {
            stack.pop()
            if(stack.length >= 1) str += s[i]
        }
    }
    return str
};


/*
( () ( () ) ) ( () )
1 21 2 32 1 0 1 21 0

Constructing a new string according to the stack length and parenthsis's value
*/