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Count Asterisks

EasyUpdated: Aug 2, 2025
Practice on:
LeetCode

Problem

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'*'ins _,excluding the _'*'between each pair of'|'.

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

Examples

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 '|'.

Solution

Method 1 – Toggle State with Counter 1

Intuition

We can keep a flag to track whether we are inside a pair of vertical bars. We only count '*' when we are outside any such pair. Every time we see a '|', we toggle the flag.

Approach

  1. Initialize a flag (inside) to False and a counter (ans) to 0.
  2. Iterate through each character in the string:
    • If the character is '|', toggle the inside flag.
    • If the character is '*' and inside is False, increment the counter.
  3. Return the counter.

Code

C++
class Solution {
public:
    int countAsterisks(string s) {
        int ans = 0, inside = 0;
        for (char c : s) {
            if (c == '|') inside ^= 1;
            else if (c == '*' && !inside) ans++;
        }
        return ans;
    }
};
Go
func countAsterisks(s string) int {
    ans, inside := 0, false
    for _, c := range s {
        if c == '|' {
            inside = !inside
        } else if c == '*' && !inside {
            ans++
        }
    }
    return ans
}
Java
class Solution {
    public int countAsterisks(String s) {
        int ans = 0;
        boolean inside = false;
        for (char c : s.toCharArray()) {
            if (c == '|') inside = !inside;
            else if (c == '*' && !inside) ans++;
        }
        return ans;
    }
}
Kotlin
class Solution {
    fun countAsterisks(s: String): Int {
        var ans = 0
        var inside = false
        for (c in s) {
            if (c == '|') inside = !inside
            else if (c == '*' && !inside) ans++
        }
        return ans
    }
}
Python
class Solution:
    def countAsterisks(self, s: str) -> int:
        ans = 0
        inside = False
        for c in s:
            if c == '|':
                inside = not inside
            elif c == '*' and not inside:
                ans += 1
        return ans
Rust
impl Solution {
    pub fn count_asterisks(s: String) -> i32 {
        let mut ans = 0;
        let mut inside = false;
        for c in s.chars() {
            if c == '|' {
                inside = !inside;
            } else if c == '*' && !inside {
                ans += 1;
            }
        }
        ans
    }
}
TypeScript
class Solution {
    countAsterisks(s: string): number {
        let ans = 0, inside = false;
        for (const c of s) {
            if (c === '|') inside = !inside;
            else if (c === '*' && !inside) ans++;
        }
        return ans;
    }
}

Complexity

  • ⏰ Time complexity: O(n), where n is the length of s, as we scan the string once.
  • 🧺 Space complexity: O(1), as only a constant amount of space is used.

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