README.md

1111. Maximum Nesting Depth of Two Valid Parentheses Strings

A string is a valid parentheses string (denoted VPS) if and only if it consists of "(" and ")" characters only, and:

• It is the empty string, or
• It can be written as AB (A concatenated with B), where A and B are VPS's, or
• It can be written as (A), where A is a VPS.

We can similarly define the nesting depth depth(S) of any VPS S as follows:

• depth("") = 0
• depth(A + B) = max(depth(A), depth(B)), where A and B are VPS's
• depth("(" + A + ")") = 1 + depth(A), where A is a VPS.

For example, "", "()()", and "()(()())" are VPS's (with nesting depths 0, 1, and 2), and ")(" and "(()" are not VPS's.

Given a VPS seq, split it into two disjoint subsequences A and B, such that A and B are VPS's (and A.length + B.length = seq.length).

Now choose any such A and B such that max(depth(A), depth(B)) is the minimum possible value.

Return an answer array (of length seq.length) that encodes such a choice of A and B: answer[i] = 0 if seq[i] is part of A, else answer[i] = 1. Note that even though multiple answers may exist, you may return any of them.

Example 1:

Input: seq = "(()())"
Output: [0,1,1,1,1,0]

Example 2:

Input: seq = "()(())()"
Output: [0,0,0,1,1,0,1,1]

Constraints:

• 1 <= seq.size <= 10000

Solutions (Rust)

1. Solution

impl Solution {
    pub fn max_depth_after_split(seq: String) -> Vec<i32> {
        let mut count0 = 0;
        let mut count1 = 0;
        let mut answer = vec![0; seq.len()];

        for (i, ch) in seq.chars().enumerate() {
            match (ch, count0 < count1) {
                ('(', true) => count0 += 1,
                (')', false) => count0 -= 1,
                (_, true) => count1 -= 1,
                (_, false) => count1 += 1,
            }

            answer[i] = ((ch == '(') ^ (count0 < count1)) as i32;
        }

        answer
    }
}