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by Sumit Chourasia | Oct 03, 2020 | Category :coding | Tags : algorithm data-structure easy leetcode stack

Remove Outermost Parentheses - Stacks - Easy - LeetCode

Remove Outermost Parentheses - Stacks - Easy - LeetCode

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

 

public class Solution {
    public string RemoveOuterParentheses(string S) {
        var sb = new StringBuilder();
        var stack = new Stack<char>();  
        var open = 0;
        for(int i=0;i<S.Length;i++){
            if(S[i]=='('){
                if(open>0){
                    sb.Append("(");
                }
                open++;             
            }
            else if(S[i]==')'){
                if(open>1){
                    sb.Append(")");
                }
                open--;
            }            
        }
        
        return sb.ToString();
    }
}

Time Complexity: O(n)

Space Complexity: O(n)