1021_Remove_Outermost_Parentheses.cpp

/*
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

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/remove-outermost-parentheses
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
*/

/* O(N) in time and O(N) in space */
#include <stack>
#include <queue>
#include <string>
using namespace std;

class Solution {
public:
    string removeOuterParentheses(string S) {
        string res;
        stack<int> s;
        queue<int> q;

        for(int i=0; i<(int)S.size(); ++i){
            if(S[i]=='('){
                s.push(i);
            }
            else{
                int j = s.top();
                s.pop();
                if(s.empty()){
                    q.push(j);
                    q.push(i);
                }
            }
        }
        for(int i=0; i<(int)S.size(); ++i){
            if(i==q.front()){
                q.pop();
            }
            else res.push_back(S[i]);
        }

        return res;
    }
};