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# 1249. Minimum Remove to Make Valid Parent
set string
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 <= 10``5`
* `s[i]` is either `'('` , `')'`, or lowercase English letter.
分析：只存index。stack存(， to\_remove存)用栈得到需要去掉的所有index，然后从s里去除掉这些Index
```
from collections import deque
class Solution:
def minRemoveToMakeValid(self, s: str) -> str:
stack = []
to_remove = set()
for i,c in enumerate(s):
if c == '(':
stack.append(i)
elif c == ')':
if stack:
stack.pop()
else:
to_remove.add(i)
to_remove.update(stack)
return ''.join(c for i,c in enumerate(s) if i not in to_remove)
```