# 1249. Minimum Remove to Make Valid Parentheses

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 <= 105`
• `s[i]` is either`'('` , `')'`, or lowercase English letter`.`

Solution:

``````class Solution(object):
def minRemoveToMakeValid(self, s):
stack, ret = [], ['']*len(s)
for i,c in enumerate(s):
if c=='(':
stack.append(i)
elif c==')':
if stack:
left=stack.pop()
ret[left] = s[left]
ret[i] = c
else:
ret[i] = c
return ''.join(ret)``````

Solution2:

``````class Solution:
def minRemoveToMakeValid(self, s):
stack, res = [], ""
for i, ch in enumerate(s):
if ch in {'(', ')'}:
if ch == ')':
if stack and stack[-1][1] == '(':
stack.pop()
continue
stack.append((i, ch))
# Whatever left in the stack are the ones to remove.
# Initialize a set for constant-time lookup.
idx_set = set([i for i, _ in stack])
return ''.join([ch for i, ch in enumerate(s) if i not in idx_set])``````

Solution3:

``````class Solution:
def minRemoveToMakeValid(self, s):
# stack = string segments for outStr, cur = current string segment
stack, cur = [], ''
for c in s:
if c == '(':
# add current string segment to stack, cos we're now at new stack
stack += [cur]
cur = ''
elif c == ')':
# since segments are only added via '(', this means that the previous segment was preceded by a '('
if stack:
cur = stack.pop() + '(' + cur + ')'
else:
cur += c

# add all the string segments together
while stack:
cur = stack.pop() + cur

return cur``````
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