1249. Minimum Remove to Make Valid Parentheses

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

Description

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 <= 105

• s[i] is either'(' , ')', or lowercase English letter.

ac

class Solution {
    public String minRemoveToMakeValid(String s) {
        // Edge cases
        if (s == null || s.length() == 0) return s;
        
        Stack<Integer> left = new Stack<>();
        Stack<Integer> right = new Stack<>();
        HashSet<Integer> toRemove = new HashSet<>();
        
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == '(') {
                left.push(i);
            } else if (s.charAt(i) == ')') {
                right.push(i);
            }
             
            if (right.size() > left.size()) {
                toRemove.add(right.pop());
            }
        }
        
        while (left.size() > right.size()) {
            toRemove.add(left.pop());
        }
        
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < s.length(); i++) {
            if (toRemove.contains(i)) continue;
            sb.append(s.charAt(i));
        }
        
        return sb.toString();
    }
}

// O(N) time, O(N) space. Parentheses->left&right. Similar to # 022-Generate Parentheses.