Remove outermost parentheses¶
Time: O(N); Space: O(1); 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 = P_1 + P_2 + … + P_k, where P_i 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:
len(S) <= 10000
S[i] is “(” or “)”
S is a valid parentheses string
[1]:
class Solution1(object):
def removeOuterParentheses(self, S):
"""
:type S: str
:rtype: str
"""
deep = 1
result, cnt = [], 0
for c in S:
if c == '(' and cnt >= deep:
result.append(c)
if c == ')' and cnt > deep:
result.append(c)
cnt += 1 if c == '(' else -1
return "".join(result)
[2]:
sol = Solution1()
s = "(()())(())"
assert sol.removeOuterParentheses(s) == "()()()"
s = "(()())(())(()(()))"
assert sol.removeOuterParentheses(s) == "()()()()(())"
s = "()()"
assert sol.removeOuterParentheses(s) == ""