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London| 26 March SDC | Jamal Laqdiem | Sprint 1 | Refactor Functions #196
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -1,14 +1,21 @@ | ||
| /** | ||
| * Finds common items between two arrays. | ||
| * | ||
| * Time Complexity: | ||
| * Space Complexity: | ||
| * Optimal Time Complexity: | ||
| * Time Complexity:O(n*m) | ||
| * Space Complexity:O(u) | ||
| * Optimal Time Complexity:O(n+m) | ||
| * | ||
| * @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))), | ||
| ]; | ||
| export const findCommonItems = (firstArray, secondArray) => { | ||
| const allowedItems = new Set(secondArray); | ||
|
|
||
| const uniqueValues = [ | ||
| ...new Set(firstArray.filter((item) => allowedItems.has(item))), | ||
| ]; | ||
| return uniqueValues; | ||
| }; | ||
|
|
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| console.log(findCommonItems([2, 5, 7], [5, 6, 7])); |
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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(n^2) | ||
| * Space Complexity:O(1) | ||
| * Optimal Time Complexity:O(n) | ||
| * | ||
| * @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 | ||
| */ | ||
| 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(); | ||
|
|
||
| for (let num of numbers) { | ||
| const complement = target - num; | ||
| if (seen.has(complement)) { | ||
| return true; | ||
| } | ||
| seen.add(num); | ||
| } | ||
| return false; | ||
| } | ||
|
|
||
| console.log(hasPairWithSum([1, 2, 3], 4)); |
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -1,36 +1,15 @@ | ||
| /** | ||
| * 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^2) | ||
| * Space Complexity:O(n) | ||
| * Optimal Time Complexity:O(n) | ||
| * | ||
| * @param {Array} inputSequence - Sequence to remove duplicates from | ||
| * @returns {Array} New sequence with duplicates removed | ||
| */ | ||
| export function removeDuplicates(inputSequence) { | ||
| const uniqueItems = []; | ||
|
|
||
| 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]); | ||
| } | ||
| } | ||
|
|
||
| return uniqueItems; | ||
| return [...new Set(inputSequence)]; | ||
| } | ||
|
|
||
| console.log(removeDuplicates([3, 3, 5, 6, 3, 7])); |
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Why use$m$ in the complexity of your algorithm but not in the complexity of the original algorithm?
How does the algorithm use$O(min(n, m))$ space?
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You are completely right about the inconsistency between old and new Time complex:
let's define n as len(first_sequence) and m as len(second_sequence), if the input sizes differ, m should be used in both.
Original Algorithm: as explained earlier the outer loop runs n times, the inner loop runs m times, and the not in list takes up to O(n) time. Therefore, time complexity of the original code is O(n^2 \cdot m). If we assume n = m, it simplifies to O(n^3).
New Algorithm: Converting the sequences to sets takes O(n) + O(m) time. The Python set intersection '&' average-case time complexity is O(\min(n, m)) as documented in this link :(https://wiki.python.org/moin/TimeComplexity). the overall optimal time complexity of O(n + m).
Space Complexity Proof for O(\min(n, m)):
the original algorithm allocates memory for the common_items list, an item is only appended to common_items if it exists in both first_sequence and second_sequence
Mathematically, the intersection of two sets cannot contain more elements than the smaller of the two sets
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How much space needed by
first_setandsecond_set?