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)