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

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)