London | 26-Jul_SDC | Shaghayegh Far| Sprint 1 | Analyse and Refactor Functions

Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js

@@ -9,21 +9,31 @@
  *   "product": 30 // 2 * 3 * 5
  * }
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Original code used two separate loops:
+ * - The first loop calculated the sum.
+ * - The second loop calculated the product.
+ *
+ * I changed the code to use one loop for both calculations.
+ * This means we do not need to go through the array twice.
+ *
+ * Time Complexity: O(n)
+ * Space Complexity: O(1)
+ * Optimal Time Complexity: O(n)
+ *
+ * Time complexity is O(n) because we need to visit every number.
+ * Space complexity is O(1) because we use a fixed number of variables.
+ * O(n) is the best possible time complexity for this task.
  *
  * @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;
+
+  // Calculate the sum and product in the same loop
   for (const num of numbers) {
+    sum += num;
     product *= num;
   }
 

Sprint-1/JavaScript/findCommonItems/findCommonItems.js

@@ -1,14 +1,39 @@
 /**
  * Finds common items between two arrays.
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * The original code used includes() to search the second array
+ * for each item in the first array. This can take a long time
+ * when the arrays are large.
+ *
+ * I changed the code to use a Set for faster searching.
+ * First, we store the items from the second array in a Set.
+ * Then, we check which items from the first array are in the Set.
+ * We use another Set to remove duplicate items from the result.
+ *
+ * Time Complexity: O(n + m)
+ * Space Complexity: O(n + m)
+ * Optimal Time Complexity: O(n + m)
+ *
+ * The original time complexity was O(n * m) because we searched
+ * the second array for every item in the first array.
+ *
+ * The new time complexity is O(n + m) on average because
+ * we create a Set and then check the items in the first array.
+ *
+ * The space complexity is O(n + m) because the Sets and the
+ * resulting array can store items from the input arrays.
+ *
+ * O(n + m) is optimal because we need to check the arrays
+ * to find their common items.
  *
  * @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 secondSet = new Set(secondArray);
+
+  const commonItems = firstArray.filter((item) => secondSet.has(item));
+
+  return [...new Set(commonItems)];
+};

Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js

@@ -1,21 +1,50 @@
 /**
  * Find if there is a pair of numbers that sum to a given target value.
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * The original code used two loops to check every possible pair
+ * of numbers. This can take a long time when the array is large.
+ *
+ * I changed the code to use a Set to remember the numbers
+ * we have already checked.
+ *
+ * For each number, we calculate the number we need to find
+ * to reach the target. If we have already seen that number,
+ * we return true.
+ *
+ * Time Complexity: O(n)
+ * Space Complexity: O(n)
+ * Optimal Time Complexity: O(n)
+ *
+ * The original time complexity was O(n^2) because we used
+ * two loops to check the pairs of numbers.
+ *
+ * The new time complexity is O(n) on average because we
+ * only need to go through the array once.
+ *
+ * The space complexity is O(n) because the Set can store
+ * up to all the numbers in the array.
+ *
+ * O(n) is the optimal time complexity for this approach
+ * because we may need to check every number in the array.
  *
  * @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 seenNumbers = new Set();
+
+  for (const num of numbers) {
+    const neededNumber = target - num;
+
+    // Check if we have already seen the number we need
+    if (seenNumbers.has(neededNumber)) {
+      return true;
     }
+
+    // Remember this number for the next checks
+    seenNumbers.add(num);
   }
+
   return false;
 }

Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs

@@ -1,36 +1,41 @@
 /**
- * Remove duplicate values from a sequence, preserving the order of the first occurrence of each value.
+ * Remove duplicate values from a sequence, preserving the order
+ * of the first occurrence of each value.
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * The original code used two loops to check whether each value
+ * was already in the result array.
+ *
+ * I changed the code to use a Set. A Set stores each value only
+ * once and keeps the order in which values were added.
+ *
+ * Time Complexity: O(n)
+ * Space Complexity: O(n)
+ * Optimal Time Complexity: O(n)
+ *
+ * The original time complexity was O(n^2) because we searched
+ * the result array for every item in the input array.
+ *
+ * The new time complexity is O(n) on average because we go
+ * through the input array once.
+ *
+ * The space complexity is O(n) because we may need to store
+ * all the unique items in the Set and the returned array.
+ *
+ * O(n) is optimal because we need to check every item in
+ * the input array to find all unique values.
  *
  * @param {Array} inputSequence - Sequence to remove duplicates from
  * @returns {Array} New sequence with duplicates removed
  */
 export function removeDuplicates(inputSequence) {
-  const uniqueItems = [];
+  // Create a Set to store unique items
+  const uniqueItems = new Set();
 
-  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]);
-    }
+  // Go through the input array and add each item to the Set
+  for (const item of inputSequence) {
+    uniqueItems.add(item);
   }
 
-  return uniqueItems;
+  // Convert the Set back into an array
+  return [...uniqueItems];
 }