> For the complete documentation index, see [llms.txt](https://soumyajit4419.gitbook.io/ds-algo/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://soumyajit4419.gitbook.io/ds-algo/stack-and-queue/applications/13.-minimum-remove-to-make-valid-parentheses.md).

# 14. 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)"
```

## Solution: 

**Approach:**\
**Keeping track of invalid parentheses using stack** \
**Removing them at end**

```csharp
class Solution
{
public:
    string minRemoveToMakeValid(string s)
    {

        stack<int> st;
        string res = "";
        int count = 0;

        for (int i = 0; i < s.length(); i++)
        {

            if (s[i] == '(')
            {
                st.push(i);
            }
            else if (s[i] == ')')
            {
                if (!st.empty())
                {
                    st.pop();
                }
                else
                {
                    s[i] = '*';
                }
            }
        }

        while (!st.empty())
        {
            s[st.top()] = '*';
            st.pop();
        }

        for (int i = 0; i < s.length(); i++)
        {
            if (s[i] != '*')
            {
                res += s[i];
            }
        }

        return res;
    }
};
```

**Time Complexity: O(n)**