1249. Minimum Remove to Make Valid Parentheses
Medium
Given a string s of ‘(‘ , ‘)’ and lowercase English characters.
Your task is to remove the minimum number of parentheses ( ‘(‘ or ‘)’, in any positions ) so that the resulting parentheses string is valid and return any valid string.
Formally, a parentheses string is valid if and only if:
It is the empty string, contains only lowercase characters, or
It can be written as AB (A concatenated with B), where A and B are valid strings, or
It can be written as (A), where A is a valid string.
Example 1:
Input: s = “lee(t(c)o)de)”
Output: “lee(t(c)o)de”
Explanation: “lee(t(co)de)” , “lee(t(c)ode)” would also be accepted.
Example 2:
Input: s = “a)b(c)d”
Output: “ab(c)d”
Example 3:
Input: s = “))((“
Output: “”
Explanation: An empty string is also valid.
Constraints:
1 <= s.length <= 105
s[i] is either'(‘ , ‘)’, or lowercase English letter.
Solution:
/**
* @param String $s
* @return String
*/
function minRemoveToMakeValid($s) {
$stack = [];
$openCount = 0;
for ($i = 0; $i < strlen($s); $i++) {
$char = $s[$i];
if ($char == ‘)’ && $openCount == 0) {
continue;
}
if ($char == ‘(‘) {
$openCount++;
} elseif ($char == ‘)’) {
$openCount–;
}
$stack[] = $char;
}
$openCount = 0;
$answer = [];
for ($i = count($stack) – 1; $i >= 0; $i–) {
$char = $stack[$i];
if ($char == ‘(‘ && $openCount == 0) {
continue;
}
if ($char == ‘)’) {
$openCount++;
} elseif ($char == ‘(‘) {
$openCount–;
}
$answer[] = $char;
}
return implode(”, array_reverse($answer));
}
}
