index.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 = 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:

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

// Solution

const removeOuterParentheses = (S) => {
    const result = []
    let balance = 0

    for (const char of S) {
        if (char === '(') {
            if (balance) result.push(char)
            balance++
        } else {
            balance--
            if (balance) result.push(char)
        }
    }

    return result.join('')
};