921. Minimum Add to Make Parentheses Valid

A parentheses string is valid if and only if:

    It is the empty string,
    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.

You are given a parentheses string s. In one move, you can insert a parenthesis at any position of the string.

    For example, if s = "()))", you can insert an opening parenthesis to be "(()))" or a closing parenthesis to be "())))".

Return the minimum number of moves required to make s valid.


Example 1:

Input: s = "())"
Output: 1

Example 2:

Input: s = "((("
Output: 3


Constraints:

    1 <= s.length <= 1000
    s[i] is either '(' or ')'.

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


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Algorithm Visualization




Code Snippet

			
package com.ideserve.sakshi.questions;

/**
 * <b>IDeserve <br>
 * <a href="https://www.youtube.com/c/IDeserve">https://www.youtube.com/c/IDeserve</a>
 * Minimum Add to Make Parentheses Valid
 *
 * @author Sakshi
 */
class Solution {
    // Consider all kind of strings like '(()', ')))((('
	public int minAddToMakeValid(String s) {
        int n = s.length();
        int openMinusClose = 0;
        int closeMinusOpen = 0;
        int i = 0;
        while (i < n) {
            if (s.charAt(i) == '(') {
                openMinusClose++;
            } else if (openMinusClose > 0 && s.charAt(i) == ')') {
                openMinusClose--;
            } else if (s.charAt(i) == ')') {
                closeMinusOpen++;
            }
            i++;
        }
        return openMinusClose + closeMinusOpen;
    }
}
		

Order of the Algorithm

Time Complexity is O(n)
Space Complexity is O(1)


Contribution

  • Sincere thanks from IDeserve community to Sakshi Kataria for compiling current post.

    Sakshi Kataria

    Ninja Programmer