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Formatted question description: https://leetcode.ca/all/1249.html

# 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.


Example 4:

Input: s = "(a(b(c)d)"
Output: "a(b(c)d)"


Constraints:

• 1 <= s.length <= 10^5
• s[i] is one of  '(' , ')' and lowercase English letters.

Related Topics:
String, Stack

## Solution 1. Scan Left to Right then Right to Left

// OJ: https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/
// Time: O(N)
// Space: O(1)
class Solution {
public:
string minRemoveToMakeValid(string s) {
int left = 0, i, j;
for (i = 0, j = 0; i < s.size(); ++i) {
if (isalpha(s[i]) || (s[i] == '(' && ++left) || (s[i] == ')' && --left >= 0)) s[j++] = s[i];
left = max(left, 0);
}
s.resize(j);
reverse(begin(s), end(s));
left = 0;
for (i = 0, j = 0; i < s.size(); ++i) {
if (isalpha(s[i]) || (s[i] == ')' && ++left) || (s[i] == '(' && --left >= 0)) s[j++] = s[i];
left = max(left, 0);
}
s.resize(j);
reverse(begin(s), end(s));
return s;
}
};


## Solution 2. Stack

// OJ: https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/
// Time: O(N)
// Space: O(N)
class Solution {
public:
string minRemoveToMakeValid(string s) {
stack<int> st;
for (int i = 0; i < s.size(); ++i) {
if (s[i] == '(') st.push(i);
else if (s[i] == ')') {
if (st.empty()) s[i] = '*';
else st.pop();
}
}
while (st.size()) {
s[st.top()] = '*';
st.pop();
}
s.erase(remove(begin(s), end(s), '*'), end(s));
return s;
}
};

• class Solution {
public String minRemoveToMakeValid(String s) {
Stack<Integer> stack = new Stack<Integer>();
StringBuffer sb = new StringBuffer();
int length = s.length();
int deleteCount = 0;
for (int i = 0; i < length; i++) {
char c = s.charAt(i);
if (Character.isLetter(c))
sb.append(c);
else if (c == '(') {
sb.append(c);
stack.push(i - deleteCount);
} else if (c == ')') {
if (!stack.isEmpty()) {
sb.append(c);
stack.pop();
} else
deleteCount++;
}
}
while (!stack.isEmpty()) {
int index = stack.pop();
sb.deleteCharAt(index);
}
return sb.toString();
}
}

############

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();
}
}

• // OJ: https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/
// Time: O(N)
// Space: O(1) if changing input string is allowed; otherwise O(N)
class Solution {
public:
string minRemoveToMakeValid(string s) {
int j = 0, left = 0;
for (int i = 0; i < s.size(); ++i) {
if (s[i] == '(') ++left;
else if (s[i] == ')') {
if (left == 0) continue;
--left;
}
s[j++] = s[i];
}
s.resize(j);
reverse(begin(s), end(s));
j = 0, left = 0;
for (int i = 0; i < s.size(); ++i) {
if (s[i] == ')') ++left;
else if (s[i] == '(') {
if (left == 0) continue;
--left;
}
s[j++] = s[i];
}
s.resize(j);
reverse(begin(s), end(s));
return s;
}
};

• 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])

############

# 1249. Minimum Remove to Make Valid Parentheses
# https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/

class Solution:
def minRemoveToMakeValid(self, s: str):
stack = []
s = list(s)

for i,c in enumerate(s):
if c == ")":
if stack:
stack.pop()
else:
s[i] = ""

elif c == "(":
stack.append(i)

while stack:
s[stack.pop()] = ""

return "".join(s)

def minRemoveToMakeValid(self, s: str):
stack = []
res = [""] * len(s)
forbidden = set()

for i,c in enumerate(s):
if c == ")":
if stack and stack[-1][1] == "(":
stack.pop()
res[i] = c
else:

elif c == "(":
res[i] = c
stack.append((i,c))

else:
res[i] = c

if len(stack) > 0:
forbidden |= set([i for i,_ in stack])

for i in forbidden:
res[i] = ""

return "".join(res)

def minRemoveToMakeValid(self, s: str):
stack = []
res = []
forbidden = set()

for i,c in enumerate(s):
if c != "(" and c != ")":
res.append(c)
else:
if c == ")":
if stack and stack[-1][1] == "(":
stack.pop()
res.append(c)
else:

else:
res.append(c)
stack.append((i,c))

if len(stack) > 0:
forbidden |= set([i for i,_ in stack])
ans = []
for i,x in enumerate(s):
if i not in forbidden:
ans.append(x)

return "".join(ans)

return "".join(res)

• 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)
}

• 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()
}
}