LeetCode 1021 Remove Outermost Parentheses C++

Easy

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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.

 

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:

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

 

https://leetcode.com/problems/remove-outermost-parentheses/

Remove Outermost Parentheses - LeetCode

 

class Solution {
public:
    string removeOuterParentheses(string s) {
        char arr[s.length()];
        string str;
        int top = -1;
        
        for(int i = 0; s[i]; i++){
            if(s[i] == '('){
                arr[++top] = s[i];
                if(top > 0) str += arr[top];
            }
            else if(s[i] == ')'){
                if(top != 0) str += s[i];
                top--;
            }
        }
        return str;
    }
};