Package g1001_1100.s1021_remove_outermost_parentheses

Class Solution

java.lang.Object
g1001_1100.s1021_remove_outermost_parentheses.Solution
public class Solution extends Object
1021 - Remove Outermost Parentheses.

Easy

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.

  • Constructor Details

    • Solution

      public Solution()

  • Method Details

    • removeOuterParentheses

      public String removeOuterParentheses(String s)