README.md

1249. Minimum Remove to Make Valid Parentheses

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.

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

• It is the empty string, contains only lowercase characters, 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.

Example 1:

Input: s = "lee(t(c)o)de)"
Output: "lee(t(c)o)de"
Explanation: "lee(t(co)de)" , "lee(t(c)ode)" would also be accepted.

Example 2:

Input: s = "a)b(c)d"
Output: "ab(c)d"

Example 3:

Input: s = "))(("
Output: ""
Explanation: An empty string is also valid.

Example 4:

Input: s = "(a(b(c)d)"
Output: "a(b(c)d)"

Constraints:

• 1 <= s.length <= 10^5
• s[i] is one of '(' , ')' and lowercase English letters.

Solutions (Rust)

1. Stack

impl Solution {
    pub fn min_remove_to_make_valid(s: String) -> String {
        let mut lp = vec![];
        let mut ret = vec![];

        for ch in s.bytes() {
            if ch != b')' || lp.len() > 0 {
                ret.push(ch);
            }
            if ch == b'(' {
                lp.push(ret.len() - 1);
            } else if ch == b')' {
                lp.pop();
            }
        }

        while let Some(i) = lp.pop() {
            ret.remove(i);
        }

        String::from_utf8(ret).unwrap()
    }
}