RemoveOutermostParentheses.js

// Source : https://leetcode.com/problems/remove-outermost-parentheses
// Author : Dean Shi
// Date   : 2019-06-08

/***************************************************************************************
 *
 * 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
 *
 ***************************************************************************************/

/**
 * @param {string} S
 * @return {string}
 */
var removeOuterParentheses = function(S) {
    let result = ''
    let count = 0
    const checker = {
        '(': () => count++ > 0,
        ')': () => count-- > 1,
    }
    for (const c of S) {
        if (checker[c]()) result += c;
    }
    return result;
};