How do you solve Count Asterisks on LeetCode?

Count Asterisks (LeetCode #2315) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Only count asterisks when not between pairs of '|'.

What is the time complexity of Count Asterisks?

The optimal solution for Count Asterisks runs in O(n) time and uses O(1) space. The complexity is O(n) because we only traverse the string once, making it efficient for larger inputs.

What is the brute force solution for Count Asterisks?

The brute force approach for Count Asterisks is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves iterating through the string and counting asterisks while checking if we are between pairs of vertical bars. If we are, we skip counting those asterisks. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Asterisks?

Count Asterisks uses the following concepts: String. The recommended patterns to study are: Two Pointers, String Manipulation.

What are the interview tips for Count Asterisks?

Clearly explain your thought process while coding. Consider edge cases, such as strings with no asterisks or only bars. Practice writing clean and efficient code.

What are common mistakes when solving Count Asterisks?

Forgetting to toggle the in_bar flag correctly. Counting asterisks even when inside a pair of bars.

#2315

Count Asterisks

Easy

StringTwo PointersString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution efficiently counts asterisks by using a single pass through the string, toggling a flag when encountering vertical bars, and counting asterisks only when not between bars.

⚙️

Algorithm

6 steps

1Step 1: Initialize a counter for asterisks and a boolean flag to track if we are inside a pair of '|'.
2Step 2: Loop through each character in the string.
3Step 3: If the character is '|', toggle the boolean flag.
4Step 4: If the character is '*', check the boolean flag; if it's false, increment the counter.
5Step 5: Continue until the end of the string.
6Step 6: Return the total count of asterisks.

solution.py9 lines

1def countAsterisks(s):
2    count = 0
3    in_bar = False
4    for char in s:
5        if char == '|':
6            in_bar = not in_bar
7        elif char == '*' and not in_bar:
8            count += 1
9    return count

ℹ

Complexity note: The complexity is O(n) because we only traverse the string once, making it efficient for larger inputs.

1Only count asterisks when not between pairs of '|'.
2Efficiently toggle a flag to track whether we are inside a pair of bars.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.