202609281244 Leetcode Count Asterisks

#project

You are given a string s, where every two consecutive vertical bars '|' are grouped into a pair. In other words, the 1st and 2nd '|' make a pair, the 3rd and 4th '|' make a pair, and so forth.

Return the number of '*' in s, excluding the* '*' between each pair of '|'.

Note that each '|' will belong to exactly one pair.

Example 1:

Input: s = "l|*e*et|c**o|*de|"
Output: 2
Explanation: The considered characters are underlined: "l|*e*et|c**o|*de|".
The characters between the first and second '|' are excluded from the answer.
Also, the characters between the third and fourth '|' are excluded from the answer.
There are 2 asterisks considered. Therefore, we return 2.

Example 2:

Input: s = "iamprogrammer"
Output: 0
Explanation: In this example, there are no asterisks in s. Therefore, we return 0.

Example 3:

Input: s = "yo|uar|e**|b|e***au|tifu|l"
Output: 5
Explanation: All 5 asterisks are outside of pairs

Constraints:

• 1 <= s.length <= 1000
• s consists of lowercase English letters, vertical bars '|', and asterisks '*'.
• s contains an even number of vertical bars '|'.

Solution

We simply walk the characters, keeping an in_pair boolean and a count number. When we come across a '|' character, we flip the boolean and if we come across a '*' we count it only if we are outside a pair (as stated in the problem).

pub fn count_asterisks(s: String) -> i32 {
    let mut in_pair = false;
    let mut count = 0;
    for c in s.chars() {
        match c {
            '|' => in_pair = !in_pair,
            '*' if !in_pair => count += 1,
            _ => {},
        }
    }
    count
}

Rust's match guards are a pretty cool way to handle this in an elegant way.

We could use bytes() instead of chars() and match on b'|' and b'*' because we're guaranteed valid ascii strings by the problem statement, but that's an unnecessary optimization for this small problem.

Another approach entirely would be to split the string into sections of '|' and step_by(2). This is slightly less performant but may read better for some people. I think the even number skipping is a bit opaque, so I'll just leave my solution as is.

pub fn count_asterisks(s: String) -> i32 {
    s.split('|')
        .step_by(2)
        .map(|seg| seg.matches('*').count())
        .sum::<usize>() as i32
}