1249. 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)"



Constraints:

    1 <= s.length <= 10^5
    s[i] is one of  '(' , ')' and lowercase English letters.
  • O(N)
    • first pass remove invalid ‘)’
    • second pass remove invalid ‘(‘

public String minRemoveToMakeValid(String s) {
    StringBuilder sb = new StringBuilder();
    int cnt = 0;
    for (char c : s.toCharArray()) {
        if (c == '(') {
            cnt++;
        } else if (c == ')') {
            if (cnt == 0) continue;
            cnt--;
        }
        sb.append(c);
    }

    for (int i = sb.length() - 1; i >= 0 && cnt > 0; i--) {
        if (sb.charAt(i) == '(') {
            sb.deleteCharAt(i);
            cnt--;
        }
    }
    return sb.toString();
}