Count Asterisks

Count Asterisks - Problem

String Easy

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: Each '|' will belong to exactly one pair.

Input & Output

Example 1 — Basic Case

$ Input: s = "l*e*et|code|*"

› Output: 3

💡 Note: The pairs are formed by 1st-2nd '|'. Outside the pair we have "l*e*et" (2 asterisks) and "*" (1 asterisk), totaling 3 asterisks.

Example 2 — Multiple Pairs

$ Input: s = "iamprogrammer"

› Output: 0

💡 Note: No vertical bars and no asterisks, so the count is 0.

Example 3 — All Outside

$ Input: s = "yo|uar|e**|awesome|"

› Output: 2

💡 Note: Pairs are 1st-2nd '|' and 3rd-4th '|'. Outside sections are "yo", "e**", and empty string after last '|'. Only "e**" contains asterisks (2 of them).

Constraints

1 ≤ s.length ≤ 1000
s consists of lowercase English letters, '*', and '|'.
s contains an even number of '|'.

Visualization

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Asked in

a Amazon 15 M Microsoft 8

The key insight is that vertical bars come in pairs, creating 'inside' and 'outside' zones. We only count asterisks in the outside zones. Best approach is single pass with toggle - flip a boolean flag on each '|' and count '*' only when flag is false. Time: O(n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Split by Pairs Method	O(n)	O(n)	Split string by '\|' pairs, then count asterisks in sections outside pairs
Single Pass with Toggle	O(n)	O(1)	Iterate through string, toggle inside/outside state on each '\|', count '*' only when outside

Split by Pairs Method — Algorithm Steps

Split the string by '|' delimiter
Iterate through segments at even indices (0, 2, 4...)
Count all asterisks in these outside segments

Visualization

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Step-by-Step Walkthrough

Split

Divide string by '|' delimiters

Index

Even indices = outside pairs

Count

Sum asterisks in even segments

Code -

solution.c — C

#include <stdio.h>
#include <string.h>

int solution(char* s) {
    int count = 0;
    int segmentIndex = 0;
    int len = strlen(s);
    
    for (int i = 0; i < len; i++) {
        if (s[i] == '|') {
            segmentIndex++;
        } else if (s[i] == '*' && segmentIndex % 2 == 0) {
            count++;
        }
    }
    
    return count;
}

int main() {
    char s[1001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    
    int result = solution(s);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Splitting string takes O(n), then counting asterisks in segments takes O(n)

✓ Linear Growth

Space Complexity

O(n)

Split array stores up to n segments in worst case

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Split by Pairs Method — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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