London|25-SDC-NOV| FATMA DEGIRMENCI | Sprint 1|Analyse and Refactor

Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js

@@ -9,21 +9,19 @@
  *   "product": 30 // 2 * 3 * 5
  * }
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
- *
+ * Time Complexity: There are two separate loops. Total operation O(n)+O(n) ---> O(n)
+ * Space Complexity:We are only using two variables sum and product --->O(1)
+ * Optimal Time Complexity:To calculate sum and product, every element must be visited at least once, so the optimal time complexity is O(n)
+ * Refactor Note: We can combine both calculations into a single loop for cleaner and more readable code
+ * 
  * @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 +1,16 @@
 /**
  * Finds common items between two arrays.
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
- *
+ * Time Complexity: We are checking two arrays. If the first array has length n and the second has length m, the time complexity is O(n * m), 
+   because filter iterates over the first array and includes scans the second array for each element.
+ * Space Complexity: A new Set is created to store unique common items, which in the worst case can be the size of the first array --> O(n).
+ * Optimal Time Complexity: By converting one array to a Set, we can check membership in O(1) for each element of the other array, reducing the time complexity to 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) => {
+  const firstArr = new Set(firstArray); //O(n)
+  const commons = secondArray.filter((el) => firstArr.has(el)); //O(m)
+  return [...new Set(commons)]; // remove duplicates
+};

Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js

@@ -1,20 +1,22 @@
 /**
  * Find if there is a pair of numbers that sum to a given target value.
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity:To using nested loop for compare elements time complexity will be ---> O(n*n)
+ * Space Complexity:We only use few variables -->O(1)
+ * Optimal Time Complexity:Iterate array only once and use a Set for O(1) lookups / total----->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 seen = 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;
-      }
+    const neededNumber = target - numbers[i];
+    if (seen.has(neededNumber)) {
+      return true;
+    } else {
+      seen.add(numbers[i]);
     }
   }
   return false;

Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs

@@ -1,36 +1,14 @@
 /**
  * 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*n) --> there is a nested loop
+ * Space Complexity:0(n)--> we are creating a array
+ * Optimal Time Complexity: we can use a new Set to remove all duplicates -->O(n)
  *
- * @param {Array} inputSequence - Sequence to remove duplicates from
+ * @param {Array} inputSequence - Sequence to remove duplicates from //[1,1,2,2,3,4,5,5]
  * @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]);
-    }
-  }
-
-  return uniqueItems;
+  const uniqueSequence = [...new Set(inputSequence)];
+  return uniqueSequence;
 }

Sprint-1/Python/calculate_sum_and_product/calculate_sum_and_product.py

@@ -12,20 +12,18 @@ def calculate_sum_and_product(input_numbers: List[int]) -> Dict[str, int]:
         "sum": 10, // 2 + 3 + 5
         "product": 30 // 2 * 3 * 5
     }
-    Time Complexity:
-    Space Complexity:
-    Optimal time complexity:
+    Time Complexity:There are two separate loops O(n)+O(n) total --->O(n)
+    Space Complexity:We have two variables sum and product O(1) total --->O(1)
+    Optimal time complexity: O(n) cannot be improved because we need to process each element at least once
+    but the code can be refactored for better readability.
     """
     # 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

@@ -8,14 +8,16 @@ def find_common_items(
 ) -> List[ItemType]:
     """
     Find common items between two arrays.
+    Time Complexity: O(n * m * k) in the worst case, where:
+    - n = length of first_sequence (outer loop)
+    - m = length of second_sequence (inner loop)
+    - k = length of common_items, because "i not in common_items" requires scanning this list
 
-    Time Complexity:
-    Space Complexity:
-    Optimal time complexity:
+    Space Complexity:In first implementation We only store the common items in a new list.--> O(n)
+
+
+    Optimal time complexity:This can be improved to O(n + m) by using a set for faster lookups.
     """
-    common_items: List[ItemType] = []
-    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
+    first_set=set(first_sequence) # O(n)
+    commons=[item for item in second_sequence if item in first_set] # O(m)
+    return list(set(commons)) # remove duplicates

Sprint-1/Python/has_pair_with_sum/has_pair_with_sum.py

@@ -7,12 +7,13 @@ 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:
-    Space Complexity:
-    Optimal time complexity:
+    Time Complexity: We have nested loop --> O(n*n)
+    Space Complexity:We have variables -->O(1)
+    Optimal time complexity:We can use only one loop -->O(n)
     """
-    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 num in numbers:
+        if target_sum - num in seen:
+            return True
+        seen.add(num)
     return False

Sprint-1/Python/remove_duplicates/remove_duplicates.py

@@ -7,19 +7,14 @@ 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: Outer loop O(n) Inner loop (worst case) O(n) total --->O(n*n)
+    Space complexity: We have an array ,unique_items, total --->O(n)
+    Optimal time complexity: O(n) using a set for fast lookups
     """
     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)
-
+    seen = set()
+    for value in values: # O(n)
+        if value not in seen: # O(1)
+            seen.add(value) # O(1)
+            unique_items.append(value) # O(1)
     return unique_items