Sheffield | 25-SDC-Nov | Hassan Osman | Sprint 1 | Analyse and Refactor Functions

Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js

@@ -9,21 +9,37 @@
  *   "product": 30 // 2 * 3 * 5
  * }
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity: O(2n) - Since we're looping through 2 separate loops
+ * Space Complexity: O(1) - only 2 variables used
+ * Optimal Time Complexity: O(n) - JS engine must read each number at least 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;
+//   for (const num of numbers) {
+//     product *= num;
+//   }
+
+//   return {
+//     sum: sum,
+//     product: 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 +1,32 @@
 /**
  * Finds common items between two arrays.
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity: O(n * m) - filter checks each element of firstArray and includes scans secondArray
+ * Space Complexity: O(n) - filter creates a new array where size grows as the input of matches grows. Same thing with set.
+ * Optimal Time Complexity: O(n + m) - each element must be examined at least once; using a Set allows O(1) lookups
  *
  * @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))),
+// ];
+
+
+
+export const findCommonItems = (firstArray, secondArray) => {
+  const arrayToSet = new Set(secondArray);
+  const uniqueCommonItems = [];
+  const checkedItems = new Set();
+
+  for (const element of firstArray) {
+    if (arrayToSet.has(element) && !checkedItems.has(element)) {
+      checkedItems.add(element);
+      uniqueCommonItems.push(element);
+    }
+
+  }  
+  return uniqueCommonItems;
+};

Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js

@@ -1,21 +1,38 @@
 /**
  * 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) - since it's a nested loop over the same array. 
+ * Space Complexity: O(1) - since there's no other array, set, object that gorws proportionally to the input size, space usage doesn't grow.
+ * Optimal Time Complexity: O(n) - each number is checked at least once after refactoring.
  *
  * @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;
+//       }
+//     }
+//   }
+//   return false;
+// }
+
+
+
 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;
-      }
+  let checkedNumbers = new Set();
+
+  for (const number of numbers) {
+    if (checkedNumbers.has(target - number)) {
+      return true;
     }
+    checkedNumbers.add(number);
   }
   return false;
+
 }
+

Sprint-1/Python/remove_duplicates/remove_duplicates.py

@@ -7,19 +7,33 @@ def remove_duplicates(values: Sequence[ItemType]) -> List[ItemType]:
     """
     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) - nested loop run once per element n times.Same for outer loop
+    Space complexity: O(n) - since there's another array to store unique items. its size grow proportionally as the number of unique inputs grow.
+    Optimal time complexity: O(n) - For an array of n elements, you must look at each element at least a once to check if it's duplicate.
     """
+    # unique_items = []
+
+    # for value in values:
+    #     is_duplicate = False
+    #     for existing in unique_items:
+    #         if value == existing:
+    #             is_duplicate = True
+    #             break
+    #     if not is_duplicate:
+    #         unique_items.append(value)
+
+    # return unique_items
+
     unique_items = []
+    checked_values = set()
 
     for value in values:
-        is_duplicate = False
-        for existing in unique_items:
-            if value == existing:
-                is_duplicate = True
-                break
-        if not is_duplicate:
+        if value not in checked_values:
             unique_items.append(value)
-
+            checked_values.add(value)
+
     return unique_items
+
+
+
+