analysis and refactoring

Author: Faithy4444

Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js

@@ -9,26 +9,21 @@
  *   "product": 30 // 2 * 3 * 5
  * }
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity:O(2n) originally have this as using 2 separate loops
+ * Space Complexity:O(1) no extra space used that rows with input
+ * Optimal Time Complexity:O(n) must at least visit each number once
  *
  * @param {Array<number>} numbers - Numbers to process
  * @returns {Object} Object containing running total and product
  */
 export function calculateSumAndProduct(numbers) {
   let sum = 0;
-  for (const num of numbers) {
-    sum += num;
-  }
-
   let product = 1;
+  //only one loop needed to calculate both sum and product
   for (const num of numbers) {
+    sum += num;
     product *= num;
   }
 
-  return {
-    sum: sum,
-    product: product,
-  };
+  return {sum, product};
 }

Sprint-1/JavaScript/findCommonItems/findCommonItems.js

@@ -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))),
-];
+
+// 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();
+
+  for (const item of firstArray) {
+    if (setB.has(item)) {
+      common.add(item);
+    }
+  }
+
+  return [...common];
+};
+

Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js

@@ -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();
+
+  for (const num of numbers) {
+    const complement = target - num;
+    if (seen.has(complement)) {
+      return true;
     }
+    seen.add(num);
   }
+
   return false;
 }

Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs

@@ -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
+ * 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 = [];
 
-  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);
     }
   }