Barcelona | Oct-2025-1 | Ifeoma Osegbo | Sprint 1 | Complexity

Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js

@@ -1,34 +1,28 @@
 /**
- * Calculate the sum and product of integers in a list
+ * Calculate the sum and product of integers in a list.
  *
- * Note: the "sum" is every number added together
- * and the "product" is every number multiplied together
- * so for example: [2, 3, 5] would return
- * {
- *   "sum": 10, // 2 + 3 + 5
- *   "product": 30 // 2 * 3 * 5
- * }
+ * Time Complexity: O(n)
+ * Space Complexity: O(1)
+ * Optimal Time Complexity: O(n)
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * The original solution used two separate loops.
+ * This version combines both calculations into a single loop,
+ * reducing the number of iterations while keeping constant memory usage.
  *
  * @param {Array<number>} numbers - Numbers to process
- * @returns {Object} Object containing running total and product
+ * @returns {Object} Object containing the sum 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,
+    sum,
+    product,
   };
 }

Sprint-1/JavaScript/findCommonItems/findCommonItems.js

@@ -1,14 +1,28 @@
 /**
- * Finds common items between two arrays.
+ * Finds unique common items between two arrays.
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity: O(n + m)
+ * Space Complexity: O(n + k)
+ * Optimal Time Complexity: O(n + m)
+ *
+ * The original solution used Array.includes(), which searches
+ * the second array for every element in the first array.
+ * This version converts the second array into a Set so lookups
+ * are much faster.
  *
  * @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);
+  const commonItems = new Set();
+
+  for (const item of firstArray) {
+    if (secondSet.has(item)) {
+      commonItems.add(item);
+    }
+  }
+
+  return [...commonItems];
+};

Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js

@@ -1,21 +1,59 @@
 /**
  * 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)
+ * Space Complexity: O(n)
+ * Optimal Time Complexity: O(n)
+ *
+ * The original solution used nested loops, resulting in O(n²) time.
+ * This version uses a Set to store previously seen numbers,
+ * allowing constant-time lookups.
+ *
+ * @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 (const number of numbers) {
+    const complement = target - number;
+
+    if (seen.has(complement)) {
+      return true;
+    }
+
+    seen.add(number);
+  }
+
+  return false;
+}/**
+ * Find if there is a pair of numbers that sum to a given target value.
+ *
+ * Time Complexity: O(n)
+ * Space Complexity: O(n)
+ * Optimal Time Complexity: O(n)
+ *
+ * The original solution used nested loops, resulting in O(n²) time.
+ * This version uses a Set to store previously seen numbers,
+ * allowing constant-time lookups.
  *
  * @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 number of numbers) {
+    const complement = target - number;
+
+    if (seen.has(complement)) {
+      return true;
     }
+
+    seen.add(number);
   }
+
   return false;
-}
+}

Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs

@@ -1,34 +1,26 @@
 /**
  * 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)
+ * Space Complexity: O(n)
+ * Optimal Time Complexity: O(n)
+ *
+ * Refactor:
+ * The original implementation used nested loops to check for duplicates,
+ * resulting in O(n²) time complexity. Using a Set allows constant-time
+ * lookups while preserving the insertion order of unique values.
  *
  * @param {Array} inputSequence - Sequence to remove duplicates from
  * @returns {Array} New sequence with duplicates removed
  */
 export function removeDuplicates(inputSequence) {
+  const seen = new Set();
   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 (const item of inputSequence) {
+    if (!seen.has(item)) {
+      seen.add(item);
+      uniqueItems.push(item);
     }
   }
 

Sprint-1/Python/calculate_sum_and_product/calculate_sum_and_product.py

@@ -5,27 +5,32 @@ def calculate_sum_and_product(input_numbers: List[int]) -> Dict[str, int]:
     """
     Calculate the sum and product of integers in a list.
 
-    Note: the sum is every number added together
-    and the product is every number multiplied together
-    so for example: [2, 3, 5] would return
-    {
-        "sum": 10, // 2 + 3 + 5
-        "product": 30 // 2 * 3 * 5
-    }
-    Time Complexity:
-    Space Complexity:
-    Optimal time complexity:
-    """
-    # Edge case: empty list
-    if not input_numbers:
-        return {"sum": 0, "product": 1}
+    Time Complexity: O(n)
+    Space Complexity: O(1)
+    Optimal Time Complexity: O(n)
+
+    Refactor:
+    The original implementation iterated through the list twice:
+    once to calculate the sum and once to calculate the product.
+    This refactor combines both calculations into a single loop,
+    reducing the number of iterations while maintaining O(n)
+    time complexity.
 
-    sum = 0
-    for current_number in input_numbers:
-        sum += current_number
+    Args:
+        input_numbers: List of integers.
 
+    Returns:
+        Dictionary containing the sum and product.
+    """
+
+    total_sum = 0
     product = 1
-    for current_number in input_numbers:
-        product *= current_number
 
-    return {"sum": sum, "product": product}
+    for number in input_numbers:
+        total_sum += number
+        product *= number
+
+    return {
+        "sum": total_sum,
+        "product": product,
+    }