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2116. Check if a Parentheses String Can Be Valid
Description
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(Aconcatenated withB), whereAandBare valid parentheses strings. - It can be written as
(A), whereAis a valid parentheses string.
You are given a parentheses string s and a string locked, both of length n. locked is a binary string consisting only of '0's and '1's. For each index i of locked,
- If
locked[i]is'1', you cannot changes[i]. - But if
locked[i]is'0', you can changes[i]to either'('or')'.
Return true if you can make s a valid parentheses string. Otherwise, return false.
Example 1:

Input: s = "))()))", locked = "010100"
Output: true
Explanation: locked[1] == '1' and locked[3] == '1', so we cannot change s[1] or s[3].
We change s[0] and s[4] to '(' while leaving s[2] and s[5] unchanged to make s valid.
Example 2:
Input: s = "()()", locked = "0000" Output: true Explanation: We do not need to make any changes because s is already valid.
Example 3:
Input: s = ")", locked = "0"
Output: false
Explanation: locked permits us to change s[0].
Changing s[0] to either '(' or ')' will not make s valid.
Constraints:
n == s.length == locked.length1 <= n <= 105s[i]is either'('or')'.locked[i]is either'0'or'1'.
Solutions
-
class Solution { public boolean canBeValid(String s, String locked) { int n = s.length(); if (n % 2 == 1) { return false; } int x = 0; for (int i = 0; i < n; ++i) { if (s.charAt(i) == '(' || locked.charAt(i) == '0') { ++x; } else if (x > 0) { --x; } else { return false; } } x = 0; for (int i = n - 1; i >= 0; --i) { if (s.charAt(i) == ')' || locked.charAt(i) == '0') { ++x; } else if (x > 0) { --x; } else { return false; } } return true; } } -
class Solution { public: bool canBeValid(string s, string locked) { int n = s.size(); if (n & 1) { return false; } int x = 0; for (int i = 0; i < n; ++i) { if (s[i] == '(' || locked[i] == '0') { ++x; } else if (x) { --x; } else { return false; } } x = 0; for (int i = n - 1; i >= 0; --i) { if (s[i] == ')' || locked[i] == '0') { ++x; } else if (x) { --x; } else { return false; } } return true; } }; -
class Solution: def canBeValid(self, s: str, locked: str) -> bool: n = len(s) if n & 1: return False x = 0 for i in range(n): if s[i] == '(' or locked[i] == '0': x += 1 elif x: x -= 1 else: return False x = 0 for i in range(n - 1, -1, -1): if s[i] == ')' or locked[i] == '0': x += 1 elif x: x -= 1 else: return False return True -
func canBeValid(s string, locked string) bool { n := len(s) if n%2 == 1 { return false } x := 0 for i := range s { if s[i] == '(' || locked[i] == '0' { x++ } else if x > 0 { x-- } else { return false } } x = 0 for i := n - 1; i >= 0; i-- { if s[i] == ')' || locked[i] == '0' { x++ } else if x > 0 { x-- } else { return false } } return true } -
function canBeValid(s: string, locked: string): boolean { const n = s.length; if (n & 1) { return false; } let x = 0; for (let i = 0; i < n; ++i) { if (s[i] === '(' || locked[i] === '0') { ++x; } else if (x > 0) { --x; } else { return false; } } x = 0; for (let i = n - 1; i >= 0; --i) { if (s[i] === ')' || locked[i] === '0') { ++x; } else if (x > 0) { --x; } else { return false; } } return true; }