Description (leetcode)

You are given a string s, where every two consecutive vertical bars '|' are grouped into a pair. In other words, the 1^st and 2^nd '|' make a pair, the 3^rd and 4^th '|' 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: The considered characters are underlined: “yo|uar|e**|b|e***au|tifu|l”. There are 5 asterisks considered. Therefore, we return 5.

Constraints:

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

submission

// we could solve this with a simple loop
// however fold is more fun
// another fun intuition is to split by '|',
// filter by index, flat_map and count '*'
impl Solution {
    pub fn count_asterisks(s: String) -> i32 {
        s.chars()
            .fold((false, 0), |(ignore, sum), ch| match ch {
                '*' if !ignore => (ignore, sum + 1),
                '|' => (!ignore, sum),
                _ => (ignore, sum),
            })
            .1
    }
}