London | ITP-MAY-25 | Surafel Workneh| Module: Data Groups | Sprint-1 #681
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…If no valid numbers, return null
…urn the middle plus
…t returns a copy of the original array
…ical element of an array
…numerical elements of an array
…ical elements of an array
SuWebOnes
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a-robson
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a-robson
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Jul 29, 2025
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Excellent. Code quality is high and you are making great progress. I've made some comments and suggestions.
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| // Clone and sort the array numerically | ||
| const sorted = [...nums].sort((a, b) => a - b); |
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Overall this is excellent. But is it necessary to clone the array here? Could we just sort it?
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| function dedupe(arr) { | ||
| if (!Array.isArray(arr)) return []; | ||
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| const seen = new Set(); | ||
| const result = []; | ||
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| for (const item of arr) { | ||
| if (!seen.has(item)) { | ||
| seen.add(item); | ||
| result.push(item); | ||
| } | ||
| } | ||
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| return result; | ||
| } | ||
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Very good. Using a Set is the right approach. However there are easier ways in Javascript to populate a Set from an array. Do we need a for loop at all?
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| function sum(elements) { | ||
| if (!Array.isArray(elements)) return 0; | ||
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| return elements.reduce((acc, val) => { | ||
| return typeof val === "number" && !isNaN(val) ? acc + val : acc; | ||
| }, 0); | ||
| } |
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Very good. However using filter() then reduce() is a more common approach to this type of problem. What are the advantages of using this pattern rather than having the 'filtering logic' within the reduce() function?
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