1249. Minimum Remove to Make Valid Parentheses

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>
 * 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. 
 *
 * @author Sakshi
 */
public class Solution {
    
    public String minRemoveToMakeValid(String s) {
        if(s.length() == 0) {
            return s;
        }

        int n = s.length();
        int extraOpenParCount = 0;
        char[] res = new char[n];
        int j = 0;
        for (int i = 0; i < n; i++) {
            char ch = s.charAt(i);
            if ((ch >= 'a' && ch <= 'z') || ch == '(') {
                res[j++] = ch;
                if (ch == '(') extraOpenParCount++;
            } else if (ch == ')') {
                // Eliminates non-matching close parentheses here itself
                if (extraOpenParCount > 0) {
                    extraOpenParCount--;
                    res[j++] = ch;
                }
            }
        }
        
        // For cases like: ()( , we come from right to left or you end up deleting first open parentheses which will be invalid string
        j--;
        StringBuilder st = new StringBuilder();
        for (int i = j; i >= 0; i--) {
            if (extraOpenParCount > 0 && res[i] == '(') {
                extraOpenParCount--;
            } else {
                st.append(res[i]);
            }
        }        
        
        // using st.reverse makes the code faster than prefixing or insert at the beginning
        return st.reverse().toString();
    }
}
		

Order of the Algorithm

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