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Add unit tests and a complexity analysis docstring for the "Count Asterisks" problem. - #47
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Pull request overview
Adds a Maven-structured solution + JUnit coverage for LeetCode 2315 (“Count Asterisks”), along with an in-code complexity analysis.
Changes:
- Introduces
CountAsterisks2315undersrc/main/java/com/leetcode/easywith a complexity docstring. - Adds JUnit 5 tests for the three canonical examples.
- Removes the legacy default-package implementation under
src/.
Reviewed changes
Copilot reviewed 3 out of 3 changed files in this pull request and generated 4 comments.
| File | Description |
|---|---|
| src/test/java/com/leetcode/easy/CountAsterisks2315Test.java | Adds unit tests for example inputs (but test method names are currently misnamed). |
| src/main/java/com/leetcode/easy/CountAsterisks2315.java | Adds solution implementation plus time/space complexity documentation. |
| src/CountAsterisks2315.java | Removes old default-package version (including main). |
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| void testTwoSum_Example1() { | ||
| String s = "l|*e*et|c**o|*de|"; |
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Test method name appears copy/pasted from TwoSum; it should reflect the behavior under test (e.g., countAsterisks) to keep test reports/readability accurate.
| void testTwoSum_Example2() { | ||
| String s = "iamprogrammer"; |
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Test method name appears copy/pasted from TwoSum; it should reflect the behavior under test (e.g., countAsterisks) to keep test reports/readability accurate.
| void testTwoSum_Example3() { | ||
| String s = "yo|uar|e**|b|e***au|tiful"; |
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Test method name appears copy/pasted from TwoSum; it should reflect the behavior under test (e.g., countAsterisks) to keep test reports/readability accurate.
| * Breakdown: | ||
| * - Split operation: s.split("\\|") iterates through the entire string - O(n) | ||
| * - Outer loop: Iterates through half of the segments - O(n/2) | ||
| * - Inner loop: Total characters checked is at most n - O(n) | ||
| * <p> | ||
| * Space complexity: O(n) | ||
| * Breakdown: | ||
| * - Array storage: String[] words stores segments - O(n) |
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The complexity breakdown mixes units: the outer loop is proportional to the number of split segments (words.length/2), not n/2 (n = string length). Consider defining a variable like m = words.length (with m <= n+1) and expressing the loop cost in terms of m to keep the analysis precise.
| * Breakdown: | |
| * - Split operation: s.split("\\|") iterates through the entire string - O(n) | |
| * - Outer loop: Iterates through half of the segments - O(n/2) | |
| * - Inner loop: Total characters checked is at most n - O(n) | |
| * <p> | |
| * Space complexity: O(n) | |
| * Breakdown: | |
| * - Array storage: String[] words stores segments - O(n) | |
| * Let n = s.length() and m = words.length (the number of split segments), where m <= n + 1. | |
| * Breakdown: | |
| * - Split operation: s.split("\\|") iterates through the entire string - O(n) | |
| * - Outer loop: Iterates through about half of the segments - O(m/2) = O(m) | |
| * - Inner loop: Across all iterations, the total number of characters checked is at most n - O(n) | |
| * Overall: O(n) | |
| * <p> | |
| * Space complexity: O(n) | |
| * - Array storage: String[] words holds m references to substrings whose total length is O(n) - O(n) |
|
@copilot open a new pull request to apply changes based on the comments in this thread |
…sks2315 Co-authored-by: kobukuro <[email protected]>
[WIP] Add unit tests and complexity analysis for Count Asterisks problem
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Pull request overview
Copilot reviewed 3 out of 3 changed files in this pull request and generated 3 comments.
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| String[] words = s.split("\\|"); | ||
| int res = 0; | ||
| for (int i = 0; i < words.length; i = i + 2) { | ||
| for (int j = 0; j < words[i].length(); j++) { |
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String[] words = s.split("\\|") produces pipe-delimited segments rather than "words". Renaming this local variable to something like segments/parts would make the loops and the complexity comment easier to follow.
| * Breakdown: | ||
| * - Split operation: s.split("\\|") iterates through the entire string - O(n) | ||
| * - Outer loop: Iterates through about half of the segments - O(m/2) = O(m) | ||
| * - Inner loop: Across all iterations, the total number of characters checked is at most n - O(n) |
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In the complexity breakdown, the statement "the total number of characters checked is at most n - O(n)" is not valid Big-O notation. Consider rephrasing to something like "at most n characters, so O(n)" to avoid the subtraction form.
| * - Inner loop: Across all iterations, the total number of characters checked is at most n - O(n) | |
| * - Inner loop: Across all iterations, the total number of characters checked is at most n characters, so O(n) |
| * Overall: O(n) | ||
| * <p> | ||
| * Space complexity: O(n) | ||
| * - Array storage: String[] words holds m references to substrings whose total length is O(n) - O(n) |
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The space complexity bullet says the substrings total length is "O(n) - O(n)", which reads like a typo and is mathematically confusing. It should just state that the total substring storage is O(n) (or clarify what is being subtracted, if anything).
| * - Array storage: String[] words holds m references to substrings whose total length is O(n) - O(n) | |
| * - Array storage: String[] words holds m references to substrings whose total length is O(n) |
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