1021. Remove Outermost Parentheses
Description
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.
Solutions
Solution 1
1 2 3 4 5 6 7 8 9 10 11 12 13 14 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 |
|