Remove Outermost Parentheses

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 = 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.

Constraints

• 1 <= s.length <= 105
• s[i] is either '(' or ')'.
• s is a valid parentheses string.

Solution

The problem Remove Outermost Parentheses can be solved by checking the number of parentheses that are not closed. If there are no parentheses open and s[i] is (, then i is the start index of a primitive parentheses string. If there is 1 parentheses open and s[j] is ), then j is the end index of a primitive parentheses string. s can be formed by removing all characters that are at the start or end index of all primitive parentheses strings.

Implementation

static const int fast_io = []()
{
    std::ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    return 0;
}();

class Solution
{
  public:
    string removeOuterParentheses(string s)
    {
        string ans;

        int count = 0;
        for (char ch : s)
        {
            if (ch == '(' && count++)
                ans += ch;
            else if (ch == ')' && --count)
                ans += ch;
        }

        return ans;
    }
};