Glasgow | SDC NOV 2025 | Katarzyna Kazimierczuk | Sprint 1 | Analyse and refactor

Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js

@@ -9,26 +9,34 @@
  *   "product": 30 // 2 * 3 * 5
  * }
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity: O(N) 2 for loops for loops are 0
+ * Space Complexity: O(1) 2 variables
+ * 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 sum = 0;
+	// for (const num of numbers) {
+	//   sum += num;
+	// }
 
-  let product = 1;
-  for (const num of numbers) {
-    product *= num;
-  }
+	// let product = 1;
+	// for (const num of numbers) {
+	//   product *= num;
+	// }
 
-  return {
-    sum: sum,
-    product: product,
-  };
+	// product and sum can be combined to one loop
+	let sum = 0;
+	let product = 1;
+
+	for (const num of numbers) {
+		sum += num;
+		product *= num;
+	}
+	return {
+		sum: sum,
+		product: product,
+	};
 }

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: Logarythmic? No Quadratic
+ *
+// nested so includes loops over each item in first arr
+// nested loops are not efficient and used filter and inlcudes
+//array.inlcudes method goes item by item
+ * Space Complexity: O(N)
+ * Optimal Time Complexity: o(N)?
  *
  * @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))),
+	const dictToCheck = {};
+	const doubled = [];
+
+	for (const item of firstArray) {
+		dictToCheck[item] = true;
+	}
+
+	for (const item of secondArray) {
+		if (dictToCheck[item]) {
+			doubled.push(item);
+			dictToCheck[item] = false;
+		}
+	}
+	return doubled;
+};

Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js

@@ -1,21 +1,37 @@
 /**
  * Find if there is a pair of numbers that sum to a given target value.
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity:Quadratic time
+ * nested loop so inefficient comparing of each i j to target
+ * Space Complexity: O(N)
+ * 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) {
-  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) {
+// we are now making a cake from cake slices
+// numbers are now slices of cake and target is a a number of slices taht makes a whole cake
+
+export function hasPairWithSum(slices, targetWholeCake) {
+	// 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 inventoryOfSlices = {};
+
+	for (const slice of slices) {
+		const missingPieces = targetWholeCake - slice;
+
+		if (inventoryOfSlices[missingPieces]) {
+			return true;
+		}
+
+		inventoryOfSlices[slice] = true; // add slice
+	}
+	return false;
 }

Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs

@@ -1,36 +1,44 @@
 /**
  * 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: Quadratic
+ * nested loop
+ * 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 = [];
+	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]);
-    }
-  }
+	// 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 itemsWeChecked = {};
+	for (const item of inputSequence) {
+		if (!itemsWeChecked[item]) {
+			uniqueItems.push(item);
+			itemsWeChecked[item] = true;
+		}
+	}
 
-  return uniqueItems;
+	return uniqueItems;
 }

Sprint-1/Python/calculate_sum_and_product/calculate_sum_and_product.py

@@ -12,20 +12,25 @@ 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)
+    Space Complexity: O(1)
+    Optimal time complexity:O(N)
     """
     # Edge case: empty list
     if not input_numbers:
         return {"sum": 0, "product": 1}
 
-    sum = 0
-    for current_number in input_numbers:
-        sum += current_number
+    # sum = 0
+    # for current_number in input_numbers:
+    #     sum += current_number
+
+    # product = 1
+    # for current_number in input_numbers:
+    #     product *= current_number
 
+    sum = 0 
     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

@@ -9,13 +9,26 @@ def find_common_items(
     """
     Find common items between two arrays.
 
-    Time Complexity:
-    Space Complexity:
-    Optimal time complexity:
+    Time Complexity: Quadratic
+    Space Complexity: ON)
+    Optimal time complexity: O(N)
     """
     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)
+    # 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
+
+    dictToCheck = {}
+
+    for item in first_sequence:
+	    	dictToCheck[item] = True
+
+    for item in second_sequence:	
+           if item in dictToCheck:
+                common_items.append(item)
+                dictToCheck.pop(item)
+
     return common_items
+

Sprint-1/Python/has_pair_with_sum/has_pair_with_sum.py

@@ -7,12 +7,23 @@ 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:Quadratic   
+    Space Complexity:O(N)
+    Optimal time complexity: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
     return False
+    # // i am using my cake exqmple from js 
+    inventory_of_slices = {}
+
+    for slice in numbers:
+        missing_piece = target_sum - slice
+        if missing_piece in inventory_of_slices:
+            return True
+
+        inventory_of_slices[slice] = True
+
+    return False

Sprint-1/Python/remove_duplicates/remove_duplicates.py

@@ -7,19 +7,28 @@ 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:
-    Space complexity:
-    Optimal time complexity:
+    Time complexity:quadratic
+    Space complexity:O(N)
+    Optimal time complexity:O(N)
     """
     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)
+    # 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
+    # return unique_items
+
+
+    itemsWeChecked = {}
+    for item in values:
+        if item not in itemsWeChecked:
+            unique_items.append(item)
+            itemsWeChecked[item] = True
+
+    return unique_items