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authorJack Sangdahl <[email protected]>2025-01-11 12:58:39 -0700
committerJack Sangdahl <[email protected]>2025-01-13 13:02:43 -0700
commit32b3c8cc7a72294298aabcedc2033b06f2a7ca5e (patch)
treeff4e5338da75dbe7e024a2d2fd3f0c8b1262e796 /check_parentheses_valid.py
parent86fae540ca9b40b8294541593b32600549c5a653 (diff)
downloadleetcode-32b3c8cc7a72294298aabcedc2033b06f2a7ca5e.tar.gz
leetcode-32b3c8cc7a72294298aabcedc2033b06f2a7ca5e.zip
2116. check if a parentheses string can be valid
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+# 2116. Check if a Parentheses String Can Be Valid
+
+"""
+a parentheses string is a non-empty string consisting only of '(' and ')'. it
+is valid if any of the following conditions is true: it is '()', it can be
+written as 'AB' where 'A' and 'B' are valid parentheses, it can be written as
+'(A)' where 'A' is a valid parentheses string. you are given a parentheses
+string 's' and a string 'locked', both of length 'n'. 'locked' is a binary
+string consistion only of '0's and '1's. for each indiex 'i' of 'locked'.
+"""
+
+
+class Solution(object):
+ def canBeValid(self, s, locked):
+ """
+ :type s: str
+ :type locked: str
+ :rtype: bool
+ """
+ if len(s) % 2 == 1:
+ return False
+ total = open = closed = 0
+ for i in range(len(s) - 1, -1, -1):
+ if locked[i] == "0":
+ total += 1
+ elif s[i] == "(":
+ open += 1
+ elif s[i] == ")":
+ closed += 1
+ if total - open + closed < 0:
+ return False
+ total = open = closed = 0
+ for i in range(len(s)):
+ if locked[i] == "0":
+ total += 1
+ elif s[i] == "(":
+ open += 1
+ elif s[i] == ")":
+ closed += 1
+ if total + open - closed < 0:
+ return False
+ return True
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.canBeValid(s="))()))", locked="010100"))
+ print(obj.canBeValid(s="()()", locked="0000"))
+ print(obj.canBeValid(s=")", locked="0"))