#2315

Easy

auto_awesomeString-driven solution strategy

LeetCode Problem Workspace

Count Asterisks

The 'Count Asterisks' problem asks you to count '*' excluding those between pairs of '|' in a string.

Problem Statement

You are given a string s containing lowercase letters, asterisks '', and vertical bars '|'. Each vertical bar in the string forms a pair, with consecutive bars representing a single pair. Your task is to count how many ' ' characters appear in the string, excluding those between any pair of '|'.

In other words, ignore the '' characters that appear between each vertical bar pair and count the ones outside. The string will always contain an even number of '|' characters, ensuring valid pairings. Return the number of ' ' outside the paired bars.

Examples

Example 1

Input: s = "l|*e*et|co|*de|"

Output: 2

The considered characters are underlined: "l|e et|co|*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

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

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 '|'.

Solution Approach

Iterating through the string

Start by iterating through the string character by character. Track whether you are inside a pair of '|' characters. If not inside a pair, count any '*' encountered.

Flagging the vertical bar pairs

To determine when you are inside or outside a pair of '|' characters, use a flag or boolean variable to toggle the state each time a '|' character is encountered.

Counting valid asterisks

Only increment the count for '' when it is found outside any pair of '|'. Skip counting ' ' if it appears between a pair of vertical bars.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(n), where n is the length of the string, as you only need to traverse the string once. The space complexity is O(1), as you use a few variables to keep track of the current state (inside or outside a '|'-pair).

What Interviewers Usually Probe

Checks whether the candidate can efficiently handle string traversal.
Assesses whether the candidate correctly identifies how to track the status of being inside or outside the '|' pair.
Evaluates the candidate's understanding of edge cases and proper exclusion of characters between '|' pairs.

Common Pitfalls or Variants

Common pitfalls

Failing to handle the toggling of the inside/outside state when encountering '|' characters.
Counting '*' characters within a '|' pair, even though they should be excluded.
Not accounting for edge cases where there are no '*' characters or when they appear in odd patterns.

Follow-up variants

Variant with a larger string where performance optimization is required.
Variant where the string includes other symbols along with '*' and '|', introducing more complexity in exclusion logic.
Variant where the '|' characters are nested, requiring more careful tracking of their pairs.

FAQ

What is the main strategy for solving 'Count Asterisks'?

The strategy is to iterate through the string while tracking whether you're inside or outside a pair of '|'. Only count '*' characters outside those pairs.

How do you track whether you're inside a pair of '|' characters?

You can use a boolean flag that toggles each time a '|' character is encountered. This helps determine whether a '*' is inside or outside the pair.

Can I use a stack to solve the 'Count Asterisks' problem?

While you could use a stack, a simpler and more efficient approach is to use a boolean flag to track whether you're inside a '|'-pair.

What is the time complexity of this problem?

The time complexity is O(n), where n is the length of the string, as the algorithm only needs to iterate through the string once.

Are there any edge cases I should consider for this problem?

Yes, make sure to handle cases where there are no '' characters, or where ' ' characters appear in unusual patterns, such as consecutive '*' outside of '|' pairs.

terminal

Solution

Solution 1: Simulation

We define an integer variable $\textit{ok}$ to indicate whether we can count when encountering `*`. Initially, $\textit{ok}=1$, meaning we can count.

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class Solution:
    def countAsterisks(self, s: str) -> int:
        ans, ok = 0, 1
        for c in s:
            if c == "*":
                ans += ok
            elif c == "|":
                ok ^= 1
        return ans

Continue Practicing

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