WestMidlands | 26-SDC-Mar | Chioma Okeke | 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)
  *
  * @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;
+
+  // Initial code looped twice through the array, i.e. 2 operations (n + n) = O(2n)operations => O(n). Limiting the loop to just one operation makes it n operation => O(n).
+  // The space complexity is constant as the variables do not grow when the function is called.
+
   for (const num of numbers) {
+    sum += num;
     product *= num;
   }
 

Sprint-1/JavaScript/findCommonItems/findCommonItems.js

@@ -1,14 +1,21 @@
 /**
  * Finds common items between two arrays.
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Initial code has this and can be refactored for optimisation
+ * Time Complexity: O(nm)
+ * Space Complexity: O(n)
+ * Optimal Time Complexity: O(n + m)
  *
  * @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) => [
+//   ...new Set(firstArray.filter((item) => secondArray.includes(item))),
+// ];
+// The initial solution had an expensive operation of .includes(item) search on the secondArray, resulting in O(nm) time complexity. The optimised code stores the secondArray in a set and allows O(1) lookup and reduce the complexity to O(n+m)
+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 +1,24 @@
 /**
  * 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(n^2)
+ * 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
  */
 export function hasPairWithSum(numbers, target) {
+  const seenNumbers = new Set();
   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;
-      }
+    // initial solution iterates twice, having outer and inner loop which gives us a time complexity of O(n**2). To optimise this, using a set to store seen numbers, then check whether the complement(target - number) for each num in the array has been seen and in the set. This gives a complexity of O(n)
+    let complement = target - numbers[i];
+
+    if (seenNumbers.has(complement)) {
+      return true;
     }
+    seenNumbers.add(numbers[i]);
   }
   return false;
 }

Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs

@@ -1,36 +1,22 @@
 /**
  * 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**2)
+ * Space Complexity: O(n)
+ * Optimal Time Complexity: O(n)
  *
  * @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]);
+  const seenItems = new Set();
+  let result = [];
+  // The initial solution was a nested loop, with a time complexity of O(n**2). This was optimised by using a set to track seen items, thereby reducing the complexity to O(n)
+  for (const item of inputSequence) {
+    if (!seenItems.has(item)) {
+      seenItems.add(item);
+      result.push(item);
     }
   }
-
-  return uniqueItems;
+  return result;
 }