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 withB
), whereA
andB
are valid strings, or - It can be written as
(A)
, whereA
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.
function minRemoveToMakeValid(s: string): string {
let left = 0;
let right = 0;
for (const c of s) {
if (c === '(') {
left++;
} else if (c === ')') {
if (right < left) {
right++;
}
}
}
let hasLeft = 0;
let res = '';
for (const c of s) {
if (c === '(') {
if (hasLeft < right) {
hasLeft++;
res += c;
}
} else if (c === ')') {
if (hasLeft != 0 && right !== 0) {
right--;
hasLeft--;
res += c;
}
} else {
res += c;
}
}
return res;
}
impl Solution {
pub fn min_remove_to_make_valid(s: String) -> String {
let bs = s.as_bytes();
let mut right = {
let mut left = 0;
let mut right = 0;
for c in bs.iter() {
match c {
&b'(' => left += 1,
&b')' if right < left => right += 1,
_ => {}
}
}
right
};
let mut has_left = 0;
let mut res = vec![];
for c in bs.iter() {
match c {
&b'(' => {
if has_left < right {
has_left += 1;
res.push(*c);
}
}
&b')' => {
if has_left != 0 && right != 0 {
right -= 1;
has_left -= 1;
res.push(*c);
}
}
_ => {
res.push(*c);
}
}
}
String::from_utf8_lossy(&res).to_string()
}
}