London | 25-SDC-Nov | Aida Eslamimoghadam | Sprint 1 | Analyse and refactor functions#141
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aydaeslami wants to merge 8 commits intoCodeYourFuture:mainfrom
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London | 25-SDC-Nov | Aida Eslamimoghadam | Sprint 1 | Analyse and refactor functions#141aydaeslami wants to merge 8 commits intoCodeYourFuture:mainfrom
aydaeslami wants to merge 8 commits intoCodeYourFuture:mainfrom
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aydaeslami
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📅 Sprint 1
cjyuan
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Mar 2, 2026
cjyuan
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Code is good.
Can you also show the time complexity of the original implementation?
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| * Time Complexity: | ||
| * Space Complexity: | ||
| * ANSWER: O(n + m) | ||
| Because the second array is converted into a Set once, and the first array is looped through once. | ||
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I think "Time Complexity" here refers to the time complexity of the original implementation, and "Optimal Time Complexity" refers to the time complexity of the refactored version. (Otherwise they will always have the same complexity)
| export const findCommonItems = (firstArray, secondArray) => { | ||
| const secondSet = new Set(secondArray); | ||
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| return [...new Set(firstArray.filter(item => secondSet.has(item)))]; |
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Note: Set object has a built-in method that can be used to find the "common items" between two sets.
Using built-in method usually means better performance and shorter code.
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| if i in second_sequence_set and i not in common_items: | ||
| common_items.append(i) |
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Can also consider using a built-in set operation for better performance and simpler code.
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Changelist
Refactored functions to reduce time complexity by removing nested loops and using sets for faster lookups, improving overall performance.