README.md

921. Minimum Add to Make Parentheses Valid

Given a string S of '(' and ')' parentheses, we add the minimum number of parentheses ( '(' or ')', and in any positions ) so that the resulting parentheses string is valid.

Formally, a parentheses string is valid if and only if:

• It is the empty string, 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.

Given a parentheses string, return the minimum number of parentheses we must add to make the resulting string valid.

Example 1:

Input: "())"
Output: 1

Example 2:

Input: "((("
Output: 3

Example 3:

Input: "()"
Output: 0

Example 4:

Input: "()))(("
Output: 4

Note:

1. S.length <= 1000
2. S only consists of '(' and ')' characters.

Solutions (Rust)

1. Remove Valid Parentheses from String

impl Solution {
    pub fn min_add_to_make_valid(s: String) -> i32 {
        let mut s = s;
        while s.contains("()") {
            s = s.replace("()", "");
        }
        s.len() as i32
    }
}

2. Remove Valid Parentheses by Stack

impl Solution {
    pub fn min_add_to_make_valid(s: String) -> i32 {
        let mut stack: Vec<char> = Vec::new();
        for ch in s.chars() {
            if ch == ')' && stack.ends_with(&['(']) {
                stack.pop();
            } else {
                stack.push(ch);
            }
        }
        stack.len() as i32
    }
}

3. Balance

impl Solution {
    pub fn min_add_to_make_valid(s: String) -> i32 {
        let mut left = 0;
        let mut right = 0;
        for ch in s.chars() {
            if ch == '(' {
                left += 1
            } else if left > 0{
                left -= 1
            } else {
                right += 1
            }
        }
        left + right
    }
}