A parentheses string is a non-empty string consisting only of '(' and ')'. It is valid if any of the following conditions is true:
().AB (A concatenated with B), where A and B are valid parentheses strings.(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,
locked[i] is '1', you cannot change s[i].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:
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'.class Solution {
public boolean canBeValid(String s, String locked) {
int n = s.length(), openCt = 0, liquidCt = 0;
if (n % 2 == 1)
return false;
for (int i = 0; i < n; i++) {
if (locked.charAt(i) == '0')
liquidCt++;
else if (s.charAt(i) == '(')
openCt++;
else {
if (openCt > 0)
openCt--;
else if (liquidCt > 0)
liquidCt--;
else
return false;
}
}
// Do the same but in reverse (to check validity of open parentheses)
openCt = 0;
liquidCt = 0;
for (int i = n - 1; i >= 0; i--) {
if (locked.charAt(i) == '0')
liquidCt++;
else if (s.charAt(i) == ')')
openCt++;
else {
if (openCt > 0)
openCt--;
else if (liquidCt > 0)
liquidCt--;
else
return false;
}
}
// System.out.println(openCt + "\t" + liquidCt);
return true;
}
}