Day 37.1.txt

1021. Remove Outermost Parentheses

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.



class Solution {
    public String removeOuterParentheses(String s) {
        int i,j=0,k=0,l=0;
        String a="";
        for(i=0;i<s.length();i++)
        {
            if(s.charAt(i)=='(')
                j++;
            else if(s.charAt(i)==')')
                k++;
            if(j==k)
            {
                a=a+s.substring(l+1,i);
                l=i+1;
            }
        }
        return a;
    }
}