1078. Remove Outermost Parentheses

• Difficulty: Easy

• Topics: Stack

• Similar Questions:

Problem:

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 "()()" + "()" = "()()()".

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

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

Solutions:

class Solution {
public:
    string removeOuterParentheses(string S) {
        string ret;
        int balance = 0;
        int left = 0;
        for (int i = 0; i < S.length(); ++i) {
            balance += (S[i] == '(' ? 1 : -1);
            if (balance == 1 && S[i] == '(') {
                left = i;
            } else if (balance == 0 && S[i] == ')') {
                ret.append(S.substr(left + 1, i - left - 1));
            }
        }

        return ret;
    }
};