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# Minimum Remove to Make Valid Parentheses

Updated: Mar 25, 2021

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.

Solution:

```class Solution {
public String minRemoveToMakeValid(String s) {
Stack<Integer> stack = new Stack<>();
StringBuilder sb = new StringBuilder();
int extraOpenBracket = 0, index;
char c;

for(int i = 0; i < s.length(); i++)
{
c = s.charAt(i);
if(c == '(')
{
sb.append(c);
stack.push(i);
}
else if(c == ')')
{
if(!stack.isEmpty())
{
sb.append(c);
stack.pop();
}
else extraOpenBracket++;

}
else sb.append(c);

}
while(!stack.isEmpty())
{
index = stack.pop();
sb.deleteCharAt(index - extraOpenBracket);
}

return sb.toString();
}
}```