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(version: 0)
Comparing performance of:
option 1 vs option 2
Created:
2 years ago
by:
Guest
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Tests:
option 1
let arr1 = ["123", "234", "345", "456", "567", "678", "789", "890"]; let arr2 = ["123", "234", "A45", "456", "567", "A78", "A89", "890"]; const result = [...arr1]; for (const item2 of arr2) { if (!result.some((item1) => item1 === item2)) { result.push(item2); } } return result;
option 2
let arr1 = ["123", "234", "345", "456", "567", "678", "789", "890"]; let arr2 = ["123", "234", "A45", "456", "567", "A78", "A89", "890"]; let SOURCE = [...arr1, ...arr2]; return [...new Set(SOURCE)];
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Result
option 1
option 2
Fastest:
N/A
Slowest:
N/A
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Autogenerated LLM Summary
(model
llama3.2:3b
, generated one year ago):
Let's dive into the provided benchmark JSON and explain what's being tested, compared, and some considerations. **Benchmark Overview** MeasureThat.net is a platform where users can create and run JavaScript microbenchmarks to compare different approaches. In this case, we have two individual test cases: "option 1" and "option 2". The test cases aim to measure the performance of duplicate detection in an array using JavaScript. The input arrays contain some duplicates and some non-duplicates. **Option 1** For "option 1", the benchmark definition is: ```javascript let arr1 = [\"123\", \"234\", \"345\", \"456\", \"567\", \"678\", \"789\", \"890\"];\r\nlet arr2 = [\"123\", \"234\", \"A45\", \"456\", \"567\", \"A78\", \"A89\", \"890\"];\r\nconst result = [...arr1];\r\n for (const item2 of arr2) {\r\n if (!result.some((item1) => item1 === item2)) {\r\n result.push(item2);\r\n }\r\n }\r\n return result; ``` This approach uses a simple loop to iterate through `arr2` and checks for duplicates using the `some()` method. If no duplicate is found, it adds the element to `result`. **Option 2** For "option 2", the benchmark definition is: ```javascript let arr1 = [\"123\", \"234\", \"345\", \"456\", \"567\", \"678\", \"789\", \"890\"];\r\nlet arr2 = [\"123\", \"234\", \"A45\", \"456\", \"567\", \"A78\", \"A89\", \"890\"];\r\nlet SOURCE = [...arr1, ...arr2];\r\n\r\n return [...new Set(SOURCE)]; ``` This approach uses the spread operator to concatenate `arr1` and `arr2`, and then converts the resulting array to a set using `Set()`. The `Set()` function automatically removes duplicates. **Comparison** The two approaches differ in their complexity: * Option 1: O(n^2) time complexity, where n is the length of `arr1`. * Option 2: O(n) time complexity, as it uses a set to remove duplicates. In general, Option 2 is more efficient for large input sizes. **Pros and Cons** Pros of Option 2: * Faster execution times * Less memory usage (set only stores unique elements) Cons of Option 2: * Limited control over duplicate detection (no explicit loop) * May not work as expected if the input array contains non-unique elements with different values Pros of Option 1: * More explicit control over duplicate detection * Works with any input array, including those with non-unique elements Cons of Option 1: * Slower execution times * Higher memory usage (creates a new array `result`) **Other Considerations** When to use each approach? * Use Option 2 when: + You need to process large input arrays and want optimal performance. + You're working with unique elements only. * Use Option 1 when: + You need more control over the duplicate detection process. + You're working with non-unique elements that require explicit handling. **Library Used** None is explicitly mentioned in this benchmark. However, if you use `Set()` or `some()`, you might be using built-in JavaScript features. **Special JS Feature/Syntax** There are no special JS features or syntax used in these test cases.
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