Is Count Asterisks easy or hard?

Count Asterisks is classified as an Easy problem on LeetCode with a high acceptance rate above 80%. The challenge mainly tests careful string traversal and state tracking rather than advanced algorithms.

Count Asterisks Python/Java solution

In Python or Java, iterate through the string and maintain a boolean flag that flips when encountering '|'. Increment a counter when the character is '*' and the flag indicates you are outside a pipe pair. The logic remains identical across languages and runs in O(n) time.

How to solve Count Asterisks in O(n)?

Traverse the string once while maintaining a boolean variable that indicates whether the current position is inside a pipe pair. Toggle the variable whenever a '|' character appears. If the variable shows you are outside a pipe section and the character is '*', increment the counter.

What is the best approach for Count Asterisks?

The boolean toggle scan is the best approach. Iterate through the string once and flip a flag whenever a pipe character '|' appears. Count '*' only when the flag indicates you are outside a pipe pair. This runs in O(n) time with O(1) extra space.

Is Count Asterisks asked at Google/Amazon/Meta?

Count Asterisks is categorized as an easy string-processing problem commonly used in coding practice and interview preparation. Variants of delimiter parsing and state tracking appear in interviews at companies like Amazon and Google, especially for testing string traversal logic.

What data structure is used in Count Asterisks?

No complex data structure is required. The solution relies on simple string traversal with a boolean state variable and a counter. The focus is on correctly handling delimiters while scanning the string sequentially.

What is the time complexity of Count Asterisks?

The optimal solution runs in O(n) time, where n is the length of the string. Each character is processed exactly once during a single left-to-right scan. Space complexity is O(1) because only a counter and a boolean flag are maintained.

Description Hints 2 Solution Video 3:31

2315. Count Asterisks

Easy83.3% acceptance1 company

You are given a string s, where every two consecutive vertical bars '|' are grouped into a pair. In other words, the 1^st and 2^nd '|' make a pair, the 3^rd and 4^th '|' make a pair, and so forth.

Return the number of '*' in s, excluding the '*' between each pair of '|'.

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

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 '|'.

Asked at 1 company

Google

Topics

String

Hints open one at a time

Hint 1

Iterate through each character, while maintaining whether we are currently between a pair of ‘|’ or not.

Hint 2

If we are not in between a pair of ‘|’ and there is a ‘*’, increment the answer by 1.

3:31

Still stuck?Read 3 approaches, or watch a walkthrough.

Open solution

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1class Solution:
2    def countAsterisks(self, s: str) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

"l|*e*et|c**o|*de|"

Input

"iamprogrammer"

Input

"yo|uar|e**|b|e***au|tifu|l"

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