Link: https://leetcode.cn/problems/minimum-remove-to-make-valid-parentheses/

Question

difficulty: mid
adj diff: 3

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.

Constraints:

	1 <= s.length <= 105
	s[i] is either'(' , ')', or lowercase English letter.

Code

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public String minRemoveToMakeValid(String s) {
    int p = 0;
    Stack<Integer> stack = new Stack<Integer>();
    int leftParenthisCount = 0;
    while (p < s.length()) {
        if (stack.isEmpty() && s.charAt(p) == ')') {
            s = s.substring(0, p) + s.substring(p + 1);
            continue; // remove a ')' and keep p un-moved
        } else if (s.charAt(p) == '(') {
            stack.push(p);
            leftParenthisCount++;
        } else if (s.charAt(p) == ')') {
            if (leftParenthisCount > 0) {
                leftParenthisCount--;
            } else {
                s = s.substring(0, p) + s.substring(p + 1);
                continue; // remove a ')' and keep p un-moved
            }
        }
        p++;
    }
    for (int i = 0; i < leftParenthisCount; i++) {
        p = stack.pop();
        s = s.substring(0, p) + s.substring(p + 1);
    }
    return s;
}