Package g2101_2200.s2116_check_if_a_parentheses_string_can_be_valid

Class Solution

java.lang.Object
g2101_2200.s2116_check_if_a_parentheses_string_can_be_valid.Solution
public class Solution extends Object
2116 - Check if a Parentheses String Can Be Valid\. Medium 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` (`A` concatenated with `B`), where `A` and `B` are valid parentheses strings. * 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 consisting only of `'0'`s and `'1'`s. For **each** index `i` of `locked`, * If `locked[i]` is `'1'`, you **cannot** change `s[i]`. * But if `locked[i]` is `'0'`, you **can** change `s[i]` to either `'('` or `')'`. Return `true` _if you can make `s` a valid parentheses string_. Otherwise, return `false`. **Example 1:** ![](https://assets.leetcode.com/uploads/2021/11/06/eg1.png) **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.length` * 1 <= n <= 105 * `s[i]` is either `'('` or `')'`. * `locked[i]` is either `'0'` or `'1'`.

  • Constructor Details

    • Solution

      public Solution()

  • Method Details

    • canBeValid

      public boolean canBeValid(String s, String locked)