London | 26-SDC-March | Zobeir Rigi | Sprint 1 | Analyse and Refactor Functions

Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js

@@ -9,21 +9,22 @@
  *   "product": 30 // 2 * 3 * 5
  * }
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
- *
+ * Time Complexity: O(n)
+ * Space Complexity: O(1)
+ * Optimal Time Complexity: O(n)
+ * 
+ * We must visit every number at least once,
+ * so O(n) is already optimal.
+
  * @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;
+
   for (const num of numbers) {
+    sum += num;
     product *= num;
   }
 
@@ -32,3 +33,10 @@ export function calculateSumAndProduct(numbers) {
     product: product,
   };
 }
+/*
+Time Complexity is O(n) because every number in the array must be
+processed once. Space Complexity is O(1) because only two variables
+ (sum and product) are stored regardless of the array size. O(n) is 
+ also the optimal complexity because every element must be examined
+  at least once to calculate the correct sum and product.
+  */

Sprint-1/JavaScript/findCommonItems/findCommonItems.js

@@ -1,14 +1,28 @@
 /**
  * Finds common items between two arrays.
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity: O(n²)
+ * Space Complexity: O(n)
+ * Optimal Time Complexity: O(n + m)
+ *
+ * Original complexity:
+ * The original solution used filter() and includes().
+ * filter() loops through every item in firstArray.
+ * For each item, includes() may need to search the entire
+ * secondArray. This results in O(n × m) time, which is
+ * O(n²) when both arrays are similar in size.
  *
- * @param {Array} firstArray - First array to compare
- * @param {Array} secondArray - Second array to compare
- * @returns {Array} Array containing unique common items
+ * Refactor:
+ * Convert secondArray into a Set. Set.has() provides
+ * constant-time lookups, so we only loop through each
+ * array once. This reduces the time complexity to O(n + m).
+ *
+ * Space Complexity:
+ * We store secondArray in a Set, which requires extra
+ * memory proportional to the size of secondArray.
  */
-export const findCommonItems = (firstArray, secondArray) => [
-  ...new Set(firstArray.filter((item) => secondArray.includes(item))),
-];
+export const findCommonItems = (firstArray, secondArray) => {
+  const secondSet = new Set(secondArray);
+
+  return [...new Set(firstArray.filter((item) => secondSet.has(item)))];
+};

Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js

@@ -1,21 +1,40 @@
 /**
  * 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)
+ * 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
+ * Original complexity:
+ * The original solution uses nested loops. For each number,
+ * it checks every remaining number in the array looking for
+ * a pair whose sum equals the target. Since each number can
+ * be compared with many other numbers, the time complexity
+ * is O(n²).
+ *
+ * Refactor:
+ * Use a Set to store numbers that have already been seen.
+ * For each number, calculate the value needed to reach the
+ * target and check whether it exists in the Set.
+ * Set lookups are O(1), so we only need one pass through
+ * the array.
+ *
+ * Refactored Complexity:
+ * Time Complexity: O(n)
+ * Space Complexity: O(n)
  */
 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 needed = target - num;
+
+    if (seen.has(needed)) {
+      return true;
     }
+
+    seen.add(num);
   }
+
   return false;
-}
+}

Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs

@@ -1,36 +1,35 @@
 /**
  * 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(n²)
+ * Space Complexity: O(n)
+ * Optimal Time Complexity: O(n)
  *
- * @param {Array} inputSequence - Sequence to remove duplicates from
- * @returns {Array} New sequence with duplicates removed
+ * Original complexity:
+ * The original solution uses nested loops. For each item in
+ * the input sequence, it searches through the uniqueItems
+ * array to check whether the value already exists. This
+ * results in O(n²) time complexity in the worst case.
+ *
+ * Refactor:
+ * Use a Set to keep track of values that have already been
+ * seen. Set.has() and Set.add() are O(1) operations, allowing
+ * us to process the sequence in a single loop.
+ *
+ * Refactored Complexity:
+ * Time Complexity: O(n)
+ * Space Complexity: O(n)
  */
 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);
     }
   }
 
   return uniqueItems;
-}
+}