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.
SOLUTION:
class Solution:
def removeOuterParentheses(self, s: str) -> str:
n = len(s)
stack = []
allowed = set(range(n))
for i in range(n):
if s[i] == '(':
stack.append(i)
else:
curr = stack.pop()
if len(stack) == 0:
allowed.remove(curr)
allowed.remove(i)
return "".join([s[i] for i in allowed])
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