LeetCode每日一题(2116. Check if a Parentheses String Can Be Valid)

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最新推荐文章于 2025-03-23 18:58:24 发布 · 231 阅读

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#leetcode #算法

给定一个括号字符串s和一个锁定字符串locked，长度均为n。如果满足以下条件之一，则认为括号字符串是有效的：它是'()'；可以写成AB（A与B连接），其中A和B都是有效的括号字符串；可以写成(A)，其中A是一个有效的括号字符串。若在locked为'0'的位置上可以改变s的字符，判断是否能将s变为有效的括号字符串。返回判断结果。

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:

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: