Count Asterisks — Easy Problem & Solution

Problem statement

s is divided into pairs of vertical bars: the 1st and 2nd '|' form a pair, the 3rd and 4th another, and so on. s always holds an even number of bars.

Return the number of '*' characters that lie outside every such pair.

Example 1

Input: s = "code|*kai*|ro**"
Output: 2
Explanation: The two stars inside the bar pair are excluded; the trailing `**` counts.

Example 2

Input: s = "codekairo"
Output: 0
Explanation: No stars at all.

Example 3

Input: s = "yo|uar|e**|b|e***au|tifu|l"
Output: 5
Explanation: The segments outside the pairs are `yo`, `e**`, `e***au` and `l`.

Constraints

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

How to solve Count Asterisks

One pass with a single inside flag. Each bar toggles it; a star counts only when the flag is off.

Approach

1. Start with inside = false and count = 0.
2. For each character: a '|' flips inside; a '*' with inside false increments count.
3. Return count.

Why it works

Because the bar count is guaranteed even, a single toggling flag is enough — there is no nesting to track and no unbalanced bar to recover from. Splitting the string on '|' and summing the stars in the even-indexed pieces is the same idea written differently.

Complexity

• Time — O(n)
• Space — O(1)

Pitfalls

• The bars themselves are never counted, only the asterisks.
• A star sitting immediately after an opening bar is inside, not outside.
• The flag must toggle on every bar, including the closing one.

Reference solution

Python

def countAsterisks(s: str) -> int:
    count = 0
    inside = False
    for c in s:
        if c == "|":
            inside = not inside
        elif c == "*" and not inside:
            count += 1
    return count

JavaScript

var countAsterisks = function(s) {
    var count = 0, inside = false;
    for (var i = 0; i < s.length; i++) {
        var c = s.charAt(i);
        if (c === "|") inside = !inside;
        else if (c === "*" && !inside) count++;
    }
    return count;
};

Also on the editorial tab: C, C#, C++, Go, Java, Kotlin, PHP, Ruby, Rust, Swift, TypeScript.

More strings problems

• Find the Original Typed String I Easy
• Rearrange Characters to Make Target String Easy
• Number of Strings That Appear as Substrings in Word Easy
• Circular Sentence Easy
• Number of Senior Citizens Easy
• Find Words Containing Character Easy