LeetCode in Kotlin
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(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'.
Solution
class Solution {
fun canBeValid(s: String, locked: String): Boolean {
if (s.isEmpty()) {
return true
}
if (s.length and 1 > 0) {
return false
}
if (locked.isEmpty()) {
return true
}
var numOfLockedClose = 0
var numOfLockedOpen = 0
for (i in s.indices) {
val countOfChars = i + 1
if (s[i] == ')' && locked[i] == '1') {
numOfLockedClose++
if (numOfLockedClose * 2 > countOfChars) {
return false
}
}
val j = s.length - 1 - i
if (s[j] == '(' && locked[j] == '1') {
numOfLockedOpen++
if (numOfLockedOpen * 2 > countOfChars) {
return false
}
}
}
return true
}
}