Minimum Remove to Make Valid Parentheses

class Solution {

    public String makeValid(String inputString) {
        
        // First pass: Remove excess closing parentheses
        StringBuilder tempBuilder = new StringBuilder();
        int countOpen = 0;
        int netBalance = 0;
        for (int index = 0; index < inputString.length(); index++) {
            char character = inputString.charAt(index);
            if (character == '(') {
                countOpen++;
                netBalance++;
            } else if (character == ')') {
                if (netBalance == 0) continue;
                netBalance--;
            }
            tempBuilder.append(character);
        }

        // Second pass: Remove excess opening parentheses
        StringBuilder finalResult = new StringBuilder();
        int keepOpen = countOpen - netBalance;
        for (int index = 0; index < tempBuilder.length(); index++) {
            char character = tempBuilder.charAt(index);
            if (character == '(') {
                keepOpen--;
                if (keepOpen < 0) continue;
            }
            finalResult.append(character);
        }

        return finalResult.toString();
    }
}

The given Java solution addresses the problem of removing the minimum number of parentheses to make a string of parentheses valid. The approach uses a two-pass method to ensure the string is valid without unnecessary parentheses while maintaining the original order of characters.

• In the first pass:

  • Traverse the input string character by character using a StringBuilder to construct an intermediate result.
  • Maintain a balance (netBalance) of open to close parentheses. When encountering an open parenthesis '(', increase both countOpen and netBalance. For a close parenthesis ')', decrease netBalance only if it's greater than zero, ensuring not to account for unmatched ')' characters.

• In the second pass:

  • Construct the final string from the intermediate string built in the first pass.
  • Use another StringBuilder, and process each character.
  • Allow only the necessary number of '(' based on the balance left from the first pass (keepOpen calculated as countOpen - netBalance).
  • Append only valid parentheses to the final result, thus ensuring all unbalanced '(' are omitted.

The resulting string returned by the method makeValid is the input string stripped of the minimum number of parentheses required to make its structure valid, fulfilling the problem's requirements efficiently.

class Solution:
    def makeValidParentheses(self, input_str: str) -> str:
        # First phase: erase extraneous ")"
        processed_chars = []
        count_balance = 0
        count_open_parens = 0
        for char in input_str:
            if char == "(":
                count_balance += 1
                count_open_parens += 1
            if char == ")":
                if count_balance == 0:
                    continue
                count_balance -= 1
            processed_chars.append(char)

        # Second phase: erase extra "(" starting from the end
        final_result = []
        opens_remaining = count_open_parens - count_balance
        for char in processed_chars:
            if char == "(":
                opens_remaining -= 1
                if opens_remaining < 0:
                    continue
            final_result.append(char)

        return "".join(final_result)

The Python code provided defines a method named makeValidParentheses within the Solution class, aiming to solve the problem of ensuring parentheses in a string are properly balanced. This is done via a two-phase process, designed to first remove extraneous closing parentheses and then to identify and remove surplus opening parentheses. Here's a breakdown of how this solution operates:

• The function accepts an input_str argument, representing the initial string containing potentially unbalanced parentheses.
• Initialization: Several variables are set up:
  • processed_chars: a list to hold characters that either pass the validity check or will be further checked in the second phase.
  • count_balance: a counter used to ensure there are no more right parentheses being closed than the left ones opened.
  • count_open_parens: a counter to track the number of opening parentheses.
• First Phase - Removing Extraneous Closing Parentheses ())
  1. Iterate through each character in the input_str.
  2. If an opening parenthesis is encountered, increment both count_balance and count_open_parens.
  3. If a closing parenthesis is encountered, check whether there is a preceding unmatched opening parenthesis (count_balance is non-zero). If not, this closing parenthesis is skipped.
  4. Add permissible characters to the processed_chars list.
• Second Phase - Adjusting Unmatched Opening Parentheses (()
  1. Initialize opens_remaining to the difference between total opening parentheses and matched closing ones.
  2. Iterate through the processed_chars.
  3. For each opening parenthesis, decrement opens_remaining. If decrementing results in a negative value, skip adding this character; it means this opening parenthesis doesn't have a matching closing one.
  4. Append all other characters to the final results list.
• Compile the Result:
  • Convert the list of valid characters (final_result) back into a string and return it.

This approach efficiently resolves the parentheses balancing by minimizing the number of passes over the string and using data structures that facilitate fast updates and checks.