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Minimum Remove to Make Valid Parentheses

Updated: Mar 25, 2021

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.

Solution:

class Solution {
    public String minRemoveToMakeValid(String s) {
        Stack<Integer> stack = new Stack<>();
		StringBuilder sb = new StringBuilder();
		int extraOpenBracket = 0, index;
		char c;

		for(int i = 0; i < s.length(); i++) 
        {
			c = s.charAt(i);
			if(c == '(') 
            {
				sb.append(c);
				stack.push(i);
			}
			else if(c == ')') 
            {
				if(!stack.isEmpty()) 
                {
					sb.append(c);
					stack.pop();
				} 
				else extraOpenBracket++;
			
			}
			else sb.append(c);

		}
		while(!stack.isEmpty()) 
        {
			index = stack.pop();
			sb.deleteCharAt(index - extraOpenBracket);
		}

		return sb.toString();
    }
}

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