Count Asterisks - Practice Coding Problems

Count Asterisks - Problem

String Easy

Imagine you have a string containing text, asterisks (*), and vertical bars (|) that act as toggle switches. Your mission is to count the asterisks, but here's the twist: some asterisks are "hidden" between pairs of vertical bars!

The Rules:

Vertical bars (|) always come in pairs: the 1st and 2nd form a pair, the 3rd and 4th form a pair, and so on
Asterisks between a pair of vertical bars are hidden and should NOT be counted
Asterisks outside any pair should be counted

Example: In the string "l|*e*et|c**o|*de|", the asterisks between the 1st-2nd bars (*e*) and between the 3rd-4th bars (*) are hidden. Only the two asterisks in c**o should be counted, giving us 2.

Think of the vertical bars as curtains that hide the asterisks inside them!

Input & Output

example_1.py — Basic Case

$ Input: s = "l|*e*et|c**o|*de|"

› Output: 2

💡 Note: The asterisks between the 1st-2nd bars (*e*) and between the 3rd-4th bars (*) are hidden. Only the two asterisks in 'c**o' are visible and counted.

example_2.py — No Hidden Stars

$ Input: s = "iamprogrammer"

› Output: 0

💡 Note: There are no asterisks in the string, so the count is 0.

example_3.py — All Stars Visible

$ Input: s = "yo|bro|***|sis|"

› Output: 3

💡 Note: All three asterisks are between the 2nd and 3rd bars (in the visible area), so all are counted.

Constraints

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

Visualization

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Understanding the Visualization

Curtains Open

We start with curtains open (visible=true). Any asterisks here are counted.

First Curtain

When we hit the first '|', the curtain closes (visible=false). Asterisks are now hidden.

Second Curtain

The next '|' opens the curtain again (visible=true). We can count asterisks again.

Continue Pattern

This pattern continues: each '|' toggles between open/closed states.

Key Takeaway

🎯 Key Insight: The optimal solution treats each vertical bar as a toggle switch. We maintain a single boolean state and flip it each time we encounter a bar. This elegant approach eliminates the need to track all bar positions and pair relationships, achieving O(n) time complexity.

Asked in

G Google 12 a Amazon 8 ⊞ Microsoft 6 f Meta 4

The optimal solution uses a simple state tracking approach. We maintain a boolean variable to track whether we're currently in a visible area (outside bar pairs) or hidden area (between bar pairs). As we iterate through the string once, we toggle the state each time we encounter a vertical bar, and count asterisks only when we're in the visible state. This achieves O(n) time complexity with O(1) space.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Check Each Asterisk)	O(n²)	O(1)	For each asterisk, check if it's between any pair of vertical bars
One-Pass State Tracking (Optimal)	O(n)	O(1)	Track visibility state while iterating through the string once

Brute Force (Check Each Asterisk) — Algorithm Steps

Iterate through each character in the string
When we find an asterisk, collect all positions of vertical bars
Check if the asterisk position falls between any pair of bars (1st-2nd, 3rd-4th, etc.)
If the asterisk is not between any pair, count it
Return the total count

Visualization

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Step-by-Step Walkthrough

Find Asterisk

Locate an asterisk at position i

Collect All Bars

Scan entire string to find all vertical bar positions

Check Pairs

For each pair of bars (1st-2nd, 3rd-4th, etc.), check if asterisk is between them

Count or Skip

If asterisk is not between any pair, increment counter

Code -

solution.c — C

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

int countAsterisks(char* s) {
    int count = 0;
    int n = strlen(s);
    
    for (int i = 0; i < n; i++) {
        if (s[i] == '*') {
            // Find all bar positions
            int bars[1000];
            int barCount = 0;
            for (int j = 0; j < n; j++) {
                if (s[j] == '|') {
                    bars[barCount++] = j;
                }
            }
            
            // Check if asterisk is between any pair
            bool isHidden = false;
            for (int k = 0; k < barCount - 1; k += 2) {
                if (bars[k] < i && i < bars[k + 1]) {
                    isHidden = true;
                    break;
                }
            }
            
            if (!isHidden) {
                count++;
            }
        }
    }
    
    return count;
}

int main() {
    char s[1001];
    scanf("%s", s);
    // Remove quotes if present
    if (s[0] == '"') {
        int len = strlen(s);
        for (int i = 0; i < len - 2; i++) {
            s[i] = s[i + 1];
        }
        s[len - 2] = '\0';
    }
    printf("%d\n", countAsterisks(s));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each asterisk, we potentially scan the entire string to find all bar positions and check pairs

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables to track positions and count

✓ Linear Space

22.3K Views

Medium Frequency

~5 min Avg. Time

892 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Check Each Asterisk) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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