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ZA | Faith Muzondo | Module-Complexity | Sprint 1 | Analyse and refactor #51
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -1,14 +1,26 @@ | ||
| /** | ||
| * Finds common items between two arrays. | ||
| * | ||
| * Time Complexity: | ||
| * Space Complexity: | ||
| * Optimal Time Complexity: | ||
| * Time Complexity:O(n+m) it build set and loop through arrays once | ||
| * Space Complexity: store second array in set | ||
| * Optimal Time Complexity:O(m+n) | ||
| * | ||
| * @param {Array} firstArray - First array to compare | ||
| * @param {Array} secondArray - Second array to compare | ||
| * @returns {Array} Array containing unique common items | ||
| */ | ||
| export const findCommonItems = (firstArray, secondArray) => [ | ||
| ...new Set(firstArray.filter((item) => secondArray.includes(item))), | ||
| ]; | ||
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| // Refactored to use a Set for faster lookups, making the code more efficient | ||
| export const findCommonItems = (firstArray, secondArray) => { | ||
| const setB = new Set(secondArray); | ||
| const common = new Set(); | ||
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| for (const item of firstArray) { | ||
| if (setB.has(item)) { | ||
| common.add(item); | ||
|
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. It's usually faster to append an element to an array than adding an item to a set even though both operations are O(1). |
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| } | ||
| } | ||
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| return [...common]; | ||
| }; | ||
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -1,21 +1,25 @@ | ||
| /** | ||
| * Find if there is a pair of numbers that sum to a given target value. | ||
| * | ||
| * Time Complexity: | ||
| * Space Complexity: | ||
| * Optimal Time Complexity: | ||
| * Time Complexity: o(n2) the function use 2 nested loop | ||
| * Space Complexity:o(1)no additional significant memory used | ||
| * Optimal Time Complexity: O(n) — in the refactored version using a Set for lookups | ||
| * | ||
| * @param {Array<number>} numbers - Array of numbers to search through | ||
| * @param {number} target - Target sum to find | ||
| * @returns {boolean} True if pair exists, false otherwise | ||
| */ | ||
| //The original version used two nested loops, meaning it checked every possible pair of numbers in the array to see if they added up to the target. This approach has a time complexity of O(n²), which becomes very slow as the list grows larger. | ||
| export function hasPairWithSum(numbers, target) { | ||
| for (let i = 0; i < numbers.length; i++) { | ||
| for (let j = i + 1; j < numbers.length; j++) { | ||
| if (numbers[i] + numbers[j] === target) { | ||
| return true; | ||
| } | ||
| const seen = new Set(); | ||
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| for (const num of numbers) { | ||
| const complement = target - num; | ||
| if (seen.has(complement)) { | ||
| return true; | ||
| } | ||
| seen.add(num); | ||
| } | ||
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| return false; | ||
| } |
| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -1,34 +1,21 @@ | ||
| /** | ||
| * Remove duplicate values from a sequence, preserving the order of the first occurrence of each value. | ||
| * | ||
| * Time Complexity: | ||
| * Space Complexity: | ||
| * Optimal Time Complexity: | ||
| * Time Complexity: O(n) — the function iterates through the input sequence once | ||
|
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. In the original code, the inner loop would iterate through |
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| * Space Complexity: O(n) — in the worst case, all elements are unique and stored in the Set and output array | ||
| * Optimal Time Complexity: O(n) — must at least visit each element once | ||
| * | ||
| * @param {Array} inputSequence - Sequence to remove duplicates from | ||
| * @returns {Array} New sequence with duplicates removed | ||
| */ | ||
| export function removeDuplicates(inputSequence) { | ||
| const seen = new Set(); | ||
| const uniqueItems = []; | ||
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| for ( | ||
| let currentIndex = 0; | ||
| currentIndex < inputSequence.length; | ||
| currentIndex++ | ||
| ) { | ||
| let isDuplicate = false; | ||
| for ( | ||
| let compareIndex = 0; | ||
| compareIndex < uniqueItems.length; | ||
| compareIndex++ | ||
| ) { | ||
| if (inputSequence[currentIndex] === uniqueItems[compareIndex]) { | ||
| isDuplicate = true; | ||
| break; | ||
| } | ||
| } | ||
| if (!isDuplicate) { | ||
| uniqueItems.push(inputSequence[currentIndex]); | ||
| for (const item of inputSequence) { | ||
| if (!seen.has(item)) { | ||
| seen.add(item); | ||
| uniqueItems.push(item); | ||
| } | ||
| } | ||
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Choose a reason for hiding this comment
The reason will be displayed to describe this comment to others. Learn more.
In complexity analysis, O(2n) is the same as O(n); the constant factor is ignored.