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

implSolution{pubfnmin_add_to_make_valid(s:String) -> i32{letmut s = s;while s.contains("()"){ s = s.replace("()","");} s.len()asi32}}

2. Remove Valid Parentheses by Stack

implSolution{pubfnmin_add_to_make_valid(s:String) -> i32{letmut stack:Vec<char> = Vec::new();for ch in s.chars(){if ch == ')' && stack.ends_with(&['(']){ stack.pop();}else{ stack.push(ch);}} stack.len()asi32}}

3. Balance

implSolution{pubfnmin_add_to_make_valid(s:String) -> i32{letmut left = 0;letmut right = 0;for ch in s.chars(){if ch == '('{ left += 1}elseif left > 0{ left -= 1}else{ right += 1}} left + right }}