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London | 26-SDC-Mar | Craig D'Silva | Sprint 1 | Analyse and Refactor Functions #170

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London | 26-SDC-Mar | Craig D'Silva | Sprint 1 | Analyse and Refactor Functions #170

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f620b8c
calculateSumAndProduct
craig-dsilva Jun 16, 2026
345b688
findCommonItems
craig-dsilva Jun 16, 2026
50abf7a
hasPairWithSum
craig-dsilva Jun 19, 2026
48e1bb3
removeDuplicates
craig-dsilva Jun 19, 2026
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8 changes: 4 additions & 4 deletions Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js
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Original file line number Diff line number Diff line change
Expand Up @@ -18,12 +18,9 @@
*/
export function calculateSumAndProduct(numbers) {
let sum = 0;
for (const num of numbers) {
sum += num;
}

let product = 1;
for (const num of numbers) {
sum += num;
product *= num;
}

Expand All @@ -32,3 +29,6 @@ export function calculateSumAndProduct(numbers) {
product: product,
};
}

// This function had 2 loops so the time complexity was O(n2).
// The updated code has time complexity of O(n) since there is only 1 loop.
2 changes: 2 additions & 0 deletions Sprint-1/JavaScript/findCommonItems/findCommonItems.js
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Expand Up @@ -12,3 +12,5 @@
export const findCommonItems = (firstArray, secondArray) => [
...new Set(firstArray.filter((item) => secondArray.includes(item))),
];

// We can't improve this as we need a nested loops are necessary for comparison
2 changes: 2 additions & 0 deletions Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js
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Expand Up @@ -19,3 +19,5 @@ export function hasPairWithSum(numbers, target) {
}
return false;
}

// In order to find the pair, it needs to do 2 loops. Nothing much can be done here to improve efficiency.
28 changes: 4 additions & 24 deletions Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs
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Expand Up @@ -9,28 +9,8 @@
* @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)];
}

// The previous code had a time complexity of O(n2) because it had 2 loops to check for duplicates.
// Set has time complexity of O(n) as it just deletes duplicates when needed and the time complexity depends on the input size.
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