Package g1001_1100.s1021_remove_outermost_parentheses

Class 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 Detail

      • Solution

        public Solution()

    • Method Detail

      • removeOuterParentheses

        public String removeOuterParentheses​(String s)