London | 26-Jul_SDC | Roman Sanaye | Sprint 1 | Analyse and Refactor Functions

Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js

@@ -9,21 +9,26 @@
  *   "product": 30 // 2 * 3 * 5
  * }
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity: O(n) - one loop through the list (the original used two
+ * loops, 2n steps, which is also O(n)).
+ *
+ * Space Complexity: O(1) - only two variables (sum, product), regardless of
+ * input size
+ *
+ * Optimal Time Complexity: O(n) - every number must be read at least once, so * we can't do better than linear time
+ *
+ * Refactor: the complexity can't be reduced, but merging the two loops into
+ * one makes the code cleaner and does half the steps.
  *
  * @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;
   }
 

Sprint-1/JavaScript/findCommonItems/findCommonItems.js

@@ -1,14 +1,41 @@
 /**
- * Finds common items between two arrays.
- *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
- *
- * @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))),
-];
+* Finds common items between two arrays.
+
+* Inefficiency: the original used secondArray.includes() inside firstArray
+filter(), which is a loop inside a loop.
+
+* Originally:
+* Time Complexity: O(n × m) - filter runs n times and each includes call can
+* scan up to m items (O(n²) if the arrays are the same size).
+
+* Space Complexity: O(n) - the filtered array and the Set hold up to n items.
+
+* Optimal Time Complexity: O(n + m) - every item in both arrays must be read at
+least once.
+
+* After refactor (the complexity was reduced from quadratic to linear time by
+using a Set for O(1) average lookups instead of includes):
+
+* Time Complexity: O(n + m) - building secondSet is O(m); the loop over
+firstArray is O(n) with O(1) average  has() checks.
+
+* Space Complexity: O(n + m) - secondSet holds up to m items; commonElements
+and the final Set hold up to n    items.
+
+* Optimal Time Complexity: O(n + m) - every item in both arrays must be read at least once, so we can't do better than linear
+* 
+* @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) => {
+  const secondSet = new Set(secondArray);
+  let commonElements = [];
+
+  for (let ele of firstArray) {
+    if (secondSet.has(ele)) {
+      commonElements.push(ele);
+    }
+  }
+  return [...new Set(commonElements)];
+};

Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js

@@ -1,21 +1,43 @@
 /**
  * Find if there is a pair of numbers that sum to a given target value.
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Inefficiency: the original used a loop inside a loop to check every
+ * possible pair of numbers.
+ *
+ * Originally:
+ * Time Complexity: O(n²) - for each number, the inner loop scans the
+ * remaining numbers, so about n * (n - 1) / 2 pair checks.
+ *
+ * Space Complexity: O(1) - only the loop counters i and j are used.
+ *
+ * Optimal Time Complexity: O(n) - every number must be read at least once,
+ * and a Set gives O(1) average lookups, so one pass is enough
+ *
+ * After Refactor (reduced from quadratic to linear time by remembering the
+ * numbers already seen in a Set instead of re-scanning the list):
+ *
+ * Time Complexity: O(n) - one loop, with O(1) average has() and add().
+ *
+ * Space Complexity: O(n) - visitedNumbers can hold up to n numbers.
+ *
+ * Optimal Time Complexity: O(n) - every number must be read at least once,
+ * so we can't do better than linear.
  *
  * @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 visitedNumbers = new Set();
+
+  for (let num of numbers) {
+    const partnerNum = target - num;
+
+    if (visitedNumbers.has(partnerNum)) {
+      return true;
     }
+    visitedNumbers.add(num);
   }
+
   return false;
 }

Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs

@@ -1,36 +1,29 @@
 /**
- * Remove duplicate values from a sequence, preserving the order of the first occurrence of each value.
- *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
- *
- * @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]);
-    }
-  }
+* Remove duplicate values from a sequence, preserving the order of the first
+* occurrence of each value.
+*
+* Originally (nested loops):
+* Time Complexity: O(n²) - the outer loop runs n times and the inner loop
+* scans uniqueItems, which can grow up to n items (about n(n - 1) / 2
+* checks in the worst case, when all items are unique).
+* 
+* Space Complexity: O(n) - uniqueItems can hold up to n items.
+* 
+* Optimal Time Complexity: O(n) - every item must be read at least once, and
+* a Set gives O(1) average lookups.
 
-  return uniqueItems;
+* After refactor ([...new Set(inputSequence)]):
+* Time Complexity: O(n) - building the Set visits each item once with O(1)
+* average inserts; spreading it into an array is also O(n).
+* 
+* Space Complexity: O(n) - the Set and the result array each hold up to n items.
+* 
+* Optimal Time Complexity: O(n) - every item must be read at least once, so we
+* can't do better than linear.
+*
+* @param {Array} inputSequence - Sequence to remove duplicates from
+* @returns {Array} New sequence with duplicates removed
+*/
+export function removeDuplicates(inputSequence) {
+  return [...new Set(inputSequence)];
 }