Remove Outermost Parentheses

Tags : leetcode, string, stack, cpp, easy

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.

Examples #

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

Example 3:

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

Constraints #

Solutions #

class Solution {
public:
    string removeOuterParentheses(string s) {
        string result = "";
        int index1 = 0;
        stack<char> st;
        for(int i=0; i<s.length(); i++ ) {
            if( s[i]=='(' ) {
                st.push('(');
            }
            else {
                st.pop();
                if(st.empty()) {
                    result += s.substr( index1+1, (i-1-index1) );
                    index1 = i+1;
                }
            }
        }
        return result;
    }
};