Description#
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 <= 105s[i] is either'(' , ')', or lowercase English letter.
Solutions#
Solution 1: Two Passes#
First, we scan from left to right and remove the extra right parentheses. Then, we scan from right to left and remove the extra left parentheses.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the string $s$.
Similar problems:
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| class Solution:
def minRemoveToMakeValid(self, s: str) -> str:
stk = []
x = 0
for c in s:
if c == ')' and x == 0:
continue
if c == '(':
x += 1
elif c == ')':
x -= 1
stk.append(c)
x = 0
ans = []
for c in stk[::-1]:
if c == '(' and x == 0:
continue
if c == ')':
x += 1
elif c == '(':
x -= 1
ans.append(c)
return ''.join(ans[::-1])
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| class Solution {
public String minRemoveToMakeValid(String s) {
Deque<Character> stk = new ArrayDeque<>();
int x = 0;
for (int i = 0; i < s.length(); ++i) {
char c = s.charAt(i);
if (c == ')' && x == 0) {
continue;
}
if (c == '(') {
++x;
} else if (c == ')') {
--x;
}
stk.push(c);
}
StringBuilder ans = new StringBuilder();
x = 0;
while (!stk.isEmpty()) {
char c = stk.pop();
if (c == '(' && x == 0) {
continue;
}
if (c == ')') {
++x;
} else if (c == '(') {
--x;
}
ans.append(c);
}
return ans.reverse().toString();
}
}
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| class Solution {
public:
string minRemoveToMakeValid(string s) {
string stk;
int x = 0;
for (char& c : s) {
if (c == ')' && x == 0) continue;
if (c == '(')
++x;
else if (c == ')')
--x;
stk.push_back(c);
}
string ans;
x = 0;
while (stk.size()) {
char c = stk.back();
stk.pop_back();
if (c == '(' && x == 0) continue;
if (c == ')')
++x;
else if (c == '(')
--x;
ans.push_back(c);
}
reverse(ans.begin(), ans.end());
return ans;
}
};
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| func minRemoveToMakeValid(s string) string {
stk := []byte{}
x := 0
for i := range s {
c := s[i]
if c == ')' && x == 0 {
continue
}
if c == '(' {
x++
} else if c == ')' {
x--
}
stk = append(stk, c)
}
ans := []byte{}
x = 0
for i := len(stk) - 1; i >= 0; i-- {
c := stk[i]
if c == '(' && x == 0 {
continue
}
if c == ')' {
x++
} else if c == '(' {
x--
}
ans = append(ans, c)
}
for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 {
ans[i], ans[j] = ans[j], ans[i]
}
return string(ans)
}
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| 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;
}
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| 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()
}
}
|
