O(n) Solution showing TLE? How is this possible?
Constraints :10^5

Question:

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.

My Solution:

class Solution {
public:
    string minRemoveToMakeValid(string s) {
        string ans;
        int i,n=s.length(),c=0;
        for(i=0;i<n;i++)
        {
            if(s[i]=='(')
            {
                c++;
                ans+=s[i];
            }
            else if(s[i]==')')
            {
                if(c)
                {
                    ans+=s[i];
                    c--;
                }
            }
            else
                ans+=s[i]; 
        }
        //cout<<c<<"\n";
        if(c)
        {
            string ans1;
            n=ans.length()-1;
            for(i=n;i>=0;i--)
            {
                if(ans[i]=='(' && c)
                    c--;
                else
                    ans1=ans[i]+ans1;
                //cout<<ans1<<"\t";
            }
            return ans1;
            
        }
        return ans;
        
        
    }
};