London | 25-SDC-Nov | Aida Eslamimoghadam | Sprint 1 | Analyse and refactor functions

Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js

@@ -10,23 +10,29 @@
  * }
  *
  * Time Complexity:
+ *      ANSWER: Because the function loops through the numbers array once,and the running time increases linearly with the size of the input.
+ *              O(n). 
  * Space Complexity:
+ *      ANSWER: O(1), Because the function only uses two extra variables (sum and product) and does not create any additional data structures based on the input size.
+ * 
+
  * Optimal Time Complexity:
+            ANSWER: This is optimal because every number in the array must be processed at least once to calculate the sum and product, so it cannot be done faster than O(n).
+            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;
+
+
   for (const num of numbers) {
+    sum += num;
     product *= num;
   }
-
   return {
     sum: sum,
     product: product,

Sprint-1/JavaScript/findCommonItems/findCommonItems.js

@@ -2,13 +2,23 @@
  * Finds common items between two arrays.
  *
  * Time Complexity:
- * Space Complexity:
+ *        ANSWER: O(n + m)
+          Because the second array is converted into a Set once, and the first array is looped through once.
+
+ * Space Complexity: 
+          ANSWER: O(n)
+          Because a Set is used to store unique common item 
+
  * Optimal Time Complexity:
+          ANSWER: O(n + m)
+          Possible by using a Set for faster lookup
  *
  * @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);
+
+  return [...new Set(firstArray.filter(item => secondSet.has(item)))];
+};

Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js

@@ -2,20 +2,34 @@
  * Find if there is a pair of numbers that sum to a given target value.
  *
  * Time Complexity:
+ *      ANSWER: O(n)
+        Because the array is looped through once.
+
  * Space Complexity:
+        ANSWER: O(n)
+        Because a Set is used to store seen numbers.
+
  * Optimal Time Complexity:
+        ANSWER: O(n)
+        Each number is processed once using a Set for lookup.
  *
  * @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 seen = new Set();
+
+  for (const num of numbers) {
+    const complement = target - num;
+
+    if (seen.has(complement)) {
+      return true;
     }
+
+    seen.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:
+ * Time Complexity: 
+ *          ANSWER:O(n)
+            Because the array is looped through once.
  * Space Complexity:
+            ANSWER:O(n)
+            Because a Set is used to track seen items.
  * Optimal Time Complexity:
+            ANSWER:O(n)
+            Each item is processed once using a Set for lookup.
  *
  * @param {Array} inputSequence - Sequence to remove duplicates from
  * @returns {Array} New sequence with duplicates removed
  */
 export function removeDuplicates(inputSequence) {
-  const uniqueItems = [];
+  const seen = new Set();
+  const result = [];
 
-  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);
+      result.push(item);
     }
   }
 
-  return uniqueItems;
+  return result;
 }

Sprint-1/Python/calculate_sum_and_product/calculate_sum_and_product.py

@@ -13,19 +13,26 @@ def calculate_sum_and_product(input_numbers: List[int]) -> Dict[str, int]:
         "product": 30 // 2 * 3 * 5
     }
     Time Complexity:
+        ANSWER: O(n)
+        Because the list is looped through once.
+
     Space Complexity:
+        ANSWER: O(1)
+        Because only two variables are used.
+
     Optimal time complexity:
+        ANSWER: O(n)
+        Each number is processed once.
+
     """
     # Edge case: empty list
     if not input_numbers:
         return {"sum": 0, "product": 1}
 
     sum = 0
-    for current_number in input_numbers:
-        sum += current_number
-
     product = 1
     for current_number in input_numbers:
+        sum += current_number
         product *= current_number
 
     return {"sum": sum, "product": product}

Sprint-1/Python/find_common_items/find_common_items.py

@@ -14,8 +14,10 @@ def find_common_items(
     Optimal time complexity:
     """
     common_items: List[ItemType] = []
+    second_sequence_set = set(second_sequence)
+
     for i in first_sequence:
-        for j in second_sequence:
-            if i == j and i not in common_items:
-                common_items.append(i)
-    return common_items
+        if i in second_sequence_set and i not in common_items:
+            common_items.append(i)
+
+    return common_items

Sprint-1/Python/has_pair_with_sum/has_pair_with_sum.py

@@ -8,11 +8,25 @@ def has_pair_with_sum(numbers: List[Number], target_sum: Number) -> bool:
     Find if there is a pair of numbers that sum to a target value.
 
     Time Complexity:
+        ANSWER: O(n)
+        Because the list is looped through once.
+
     Space Complexity:
+        ANSWER: O(n)
+        Because a set is used to store seen numbers.
+
     Optimal time complexity:
+        ANSWER: O(n)
+        Each number is processed once using a set for faster lookup.
+
+
     """
-    for i in range(len(numbers)):
-        for j in range(i + 1, len(numbers)):
-            if numbers[i] + numbers[j] == target_sum:
-                return True
+    seen = set()
+
+    for number in numbers:
+        complement = target_sum - number
+        if complement in seen:
+            return True
+        seen.add(number)
+
     return False

Sprint-1/Python/remove_duplicates/remove_duplicates.py

@@ -8,18 +8,28 @@ 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:
+        ANSWER: O(n)
+        Because the sequence is looped through once.
+
     Space complexity:
+        ANSWER: O(n)
+        Because a set is used to track seen value
+
+
     Optimal time complexity:
+        ANSWER: O(n)
+        Each value is processed once using a set for faster lookup
+
     """
     unique_items = []
+    seen = 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 seen:
+            seen.add(value) 
+
             unique_items.append(value)
+
 
     return unique_items