Toggle navigation
MeasureThat.net
Create a benchmark
Tools
Feedback
FAQ
Register
Log In
Jewels and Stones
(version: 0)
Comparing performance of:
JS-Set vs JS-Array
Created:
4 years ago
by:
Guest
Jump to the latest result
Tests:
JS-Set
var numJewelsInStones = function (J, S) { const set = new Set(J.split("")); let count = 0; for (let i = 0; i < S.length; i++) { if (set.has(S[i])) { count++; } } return count; }; numJewelsInStones('aA', 'AaaaAAaaaaAaAAAAaaaaaAabbsdfdffojeworwejoijovweojoweijoewjfwejfowejfowejofjeworjwejtweojweojfweojfwepfpoewjfp')
JS-Array
var numJewelsInStones = function(J, S) { if (!J || !S) return 0; let count = 0; for (const c of S) { if (J.includes(c)) count++; } return count; }; numJewelsInStones('aA', 'AaaaAAaaaaAaAAAAaaaaaAabbsdfdffojeworwejoijovweojoweijoewjfwejfowejfowejofjeworjwejtweojweojfweojfwepfpoewjfp')
Rendered benchmark preparation results:
Suite status:
<idle, ready to run>
Run tests (2)
Previous results
Fork
Test case name
Result
JS-Set
JS-Array
Fastest:
N/A
Slowest:
N/A
Latest run results:
No previous run results
This benchmark does not have any results yet. Be the first one
to run it!
Autogenerated LLM Summary
(model
llama3.2:3b
, generated one year ago):
Let's break down the provided benchmark and explain what's being tested. **Benchmark Definition** The benchmark is defined as a JavaScript function that calculates the number of jewels in a given set of stones. The input parameters are two strings: `J` (the set of jewels) and `S` (the set of stones). The goal is to count the number of jewels in the set of stones. **Test Cases** There are two test cases: 1. **JS-Set**: This test case uses a Set data structure to store the jewels, which automatically removes duplicates. It iterates through each character in the set of stones and checks if it's present in the set of jewels using the `has()` method. If found, it increments the count. 2. **JS-Array**: This test case uses an array to represent the set of jewels. It iterates through each character in the set of stones and checks if it's present in the array using the `includes()` method. If found, it increments the count. **Options Compared** The two test cases compare the performance of using a Set data structure versus an array to represent the set of jewels. **Pros and Cons** * **JS-Set (using Set)**: + Pros: - Efficient lookup time (~O(1) on average) - Removes duplicates automatically + Cons: - Requires more memory for storing the set - May be slower in certain scenarios due to the overhead of creating and managing sets * **JS-Array (using Array)**: + Pros: - Less memory usage compared to Set - Can be faster in certain scenarios due to the efficiency of array lookups + Cons: - May have duplicate elements, which can slow down lookup time **Library/Utility** There is no specific library used in this benchmark. However, JavaScript provides built-in data structures and methods that make it easy to work with sets (e.g., `Set`, `has()`) and arrays (e.g., `Array.prototype.includes()`). **Special JS Feature/Syntax** The benchmark uses the following special feature/syntax: * `const set = new Set(J.split(""));`: Creates a new Set object from an array of strings, which splits the input string `J` into individual characters using the `split()` method. **Other Alternatives** If you're interested in exploring alternative approaches, here are some options: * Using a Trie data structure to store the set of jewels would likely provide faster lookup times compared to both Set and Array. * Utilizing a more efficient algorithm for counting duplicates, such as using a hash table or a frequency map, could potentially outperform both Set and Array. Keep in mind that these alternative approaches might come with additional overheads, complexity, or memory usage.
Related benchmarks:
indexof vs hash
Looping variable vs elements
IndexOf vs Includes vs hash vs set
Hash vs insert
sdfdffjlaksjfskjf
Comments
Confirm delete:
Do you really want to delete benchmark?