London | 25-SDC-July | Fatma Arslantas | Sprint 1 | Analyse and Refactor Functions

Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js

@@ -9,21 +9,21 @@
  *   "product": 30 // 2 * 3 * 5
  * }
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity: O(n) - We go through the list only once. The previous code had two loops (2n), but now we do everything in one loop (n).
+ * Space Complexity: O(1) - We represent the result with just two variables (sum and product), so we don't use extra memory.
+ * Optimal Time Complexity: O(n) - We have to look at every number at least once, so we can't be faster than 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;
+
+  // Refactor: Instead of looping twice, I calculate both sum and product in the same loop.
+  // This is better because we only loop through the list once.
   for (const num of numbers) {
+    sum += num;
     product *= num;
   }
 

Sprint-1/JavaScript/findCommonItems/findCommonItems.js

@@ -1,14 +1,31 @@
 /**
  * Finds common items between two arrays.
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity: O(n + m)
+ * The old code used .includes() inside .filter(), which meant a loop inside a loop (O(n^2)).
+ * By using a 'Set', we can check if an item exists instantly (O(1)).
+ *
+ * Space Complexity: O(n) - We create a Set to store the items from the first array.
+ *
+ * Optimal Time Complexity: O(n) - We must look at every item at least once to compare them.
  *
  * @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) => {
+  // Refactor: I put the first array into a Set for fast checking.
+  const itemsInFirst = new Set(firstArray);
+  const commonItems = new Set();
+
+  for (const item of secondArray) {
+    // Checking 'itemsInFirst.has(item)' is very fast O(1).
+    // The old 'includes()' was slow because it searched the whole list O(n).
+    if (itemsInFirst.has(item)) {
+      commonItems.add(item);
+    }
+  }
+
+  // Convert the Set back to an array to return it
+  return [...commonItems];
+};

Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js

@@ -1,21 +1,35 @@
 /**
  * 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)
+ * The old code used nested loops (loop inside a loop) which is slow O(n^2).
+ * I used a Set to remember the numbers we have seen. This lets us find the answer in one loop O(n).
+ * 
+ * Space Complexity: O(n) - We use a Set to store the numbers we have visited.
+ * Optimal Time Complexity: O(n) - We need to check the numbers at least once to find the pair.
  *
  * @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;
-      }
+  // Refactor: I used a Set to store numbers we have already seen.
+  const seenNumbers = new Set();
+
+  for (const num of numbers) {
+    // Calculate what number we need to reach the target.
+    // Example: If target is 10 and num is 3, we need 7.
+    const match = target - num;
+
+    // Check if the number we need is already in our Set.
+    // .has() is very fast O(1).
+    if (seenNumbers.has(match)) {
+      return true;
     }
+
+    // Add the current number to the Set so we can use it later.
+    seenNumbers.add(num);
   }
+
   return false;
 }

Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs

@@ -1,36 +1,18 @@
 /**
  * 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)
+ * The old code used a nested loop (checking the result array for every item), which is O(n^2).
+ * A Set automatically removes duplicates in one go. Converting Array -> Set -> Array takes linear time O(n).
+ * 
+ * Space Complexity: O(n) - We create a new Set/Array to store the unique items.
+ * Optimal Time Complexity: O(n) - We must look at every item at least once to see what it is.
  *
  * @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]);
-    }
-  }
-
-  return uniqueItems;
+  // Refactor: I used a Set to remove duplicates automatically.
+  // It creates a Set (removes duplicates) and spreads it back into an array.
+  return [...new Set(inputSequence)];
 }

Sprint-1/Python/calculate_sum_and_product/calculate_sum_and_product.py

@@ -12,20 +12,20 @@ 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: O(n)
+    The previous code iterated through the list twice (2n).
+    I combined them into a single loop, so we iterate only once (n).
+
+    Space Complexity: O(1) - We only use two variables (running_sum and running_product) regardless of the list size.
+    Optimal time complexity: O(n) - We must visit every number at least 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
+    running_sum = 0
+    running_product = 1
 
-    product = 1
-    for current_number in input_numbers:
-        product *= current_number
+    # Refactor: Calculate both sum and product in one loop.
+    for number in input_numbers:
+        running_sum += number
+        running_product *= number
 
-    return {"sum": sum, "product": product}
+    return {"sum": running_sum, "product": running_product}

Sprint-1/Python/find_common_items/find_common_items.py

@@ -9,13 +9,14 @@ def find_common_items(
     """
     Find common items between two arrays.
 
-    Time Complexity:
-    Space Complexity:
-    Optimal time complexity:
+    Time Complexity: O(n)
+    The old code used nested loops (loop inside a loop), which is slow O(n^2).
+    I converted the lists to Sets. Finding the intersection of two sets is linear time O(n).
+
+    Space Complexity: O(n) - We create a new set to store the items for comparison.
+    Optimal time complexity: O(n) - We must read the input lists at least once to compare them.
     """
-    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
+
+    # Refactor: I used Python Sets.
+    # set(first_sequence) & set(second_sequence) finds common items instantly.
+    return list(set(first_sequence) & set(second_sequence))

Sprint-1/Python/has_pair_with_sum/has_pair_with_sum.py

@@ -7,12 +7,25 @@ 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:
-    Space Complexity:
-    Optimal time complexity:
+    Time Complexity: O(n)
+    The previous code used nested loops (loop inside a loop), which is slow O(n^2).
+    I used a Set to remember the numbers we have seen. This allows us to find the pair in one pass O(n).
+
+    Space Complexity: O(n) - We store the visited numbers in a Set.
+    Optimal time complexity: O(n) - We have to check each number at least once.
     """
-    for i in range(len(numbers)):
-        for j in range(i + 1, len(numbers)):
-            if numbers[i] + numbers[j] == target_sum:
-                return True
+    # Refactor: I used a Set to store numbers we have already seen.
+    seen_numbers = set()
+
+    for num in numbers:
+        # Calculate the number we need to reach the target.
+        match = target_sum - num
+
+        # Check if the needed number is already in our set.
+        if match in seen_numbers:
+            return True
+
+        # Add the current number to the set so we can use it later.
+        seen_numbers.add(num)
+
     return False

Sprint-1/Python/remove_duplicates/remove_duplicates.py

@@ -7,19 +7,24 @@ 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(n)
+    The old code checked the list repeatedly, which is O(n^2).
+    I used a 'Set' to remember seen items for O(1) lookups, while appending to a list to keep the order.
+
+    Space Complexity: O(n) - We use a Set to store seen items and a List for the result.
+
+    Optimal time complexity: O(n) - We must iterate through the input at least once.
     """
+
+    # Refactor: I used a Set to track items we have seen because checking a Set is O(1).
+    # I also used a List to build the result so we keep the original order.
+    seen = set()
     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:
+        # Check if we have seen this value before.
+        if value not in seen:
             unique_items.append(value)
+            seen.add(value)
 
     return unique_items