London | Nadika Zavodovska | Module-Complexity | Sprint 1 | Analyse and refactor

Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js

@@ -9,26 +9,45 @@
  *   "product": 30 // 2 * 3 * 5
  * }
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity: O(n) original has two passes, optimised uses one pass
+ * Space Complexity: O(1) for both
+ * Optimal Time Complexity: O(n) linear time is the best possible
  *
  * @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;
-  }
+    // We went through the list two times, first to add the numbers, then to multiply them.
+    // But we can do both in one loop. This makes the code a bit simpler and faster.
+
+    // 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,
-  };
+    // return {
+    //     sum: sum,
+    //     product: product,
+    // };
+
+    // My solution
+
+    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 +1,33 @@
 /**
  * Finds common items between two arrays.
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity: O(m + n) — build set and loop through arrays once
+ * Space Complexity: O(n) — store second array in a set
+ * Optimal Time Complexity: O(m + 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))),
+// ];
+
+// My solution. Refactored to use a Set for faster lookups, making the code more efficient
+
+export const findCommonItems = (firstArray, secondArray) => {
+    // Turn secondArray into a Set to quickly check if items exist
+    const secondSet = new Set(secondArray);
+    // Create a Set to keep track of common items without duplicates
+    const commonItemsSet = new Set();
+
+    // Go through each item in firstArray
+    for (const element of firstArray) {
+        // If the item is found in secondSet, add it to commonItemsSet
+        if (secondSet.has(element)) {
+            commonItemsSet.add(element);
+        }
+    }
+    // Change the Set of common items back into an array to return
+    return [...commonItemsSet];
+};

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: O(n²) original (nested loops), O(n) refactored (using Set)
+ * Space Complexity: O(1) original, O(n) refactored (Set)
+ * Optimal Time Complexity: O(n) — in the refactored version using a Set for 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;
+//       }
+//     }
+//   }
+//   return false;
+// }
+
+// My solution -  Refactored to use a Set for faster lookup, reducing time from O(n²) to O(n) 
 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 previousNumbers = new Set();
+
+    for (const num of numbers) {
+        const neededValue = target - num;
+        if (previousNumbers.has(neededValue)) {
+            return true;
+        }
+        previousNumbers.add(num);
     }
-  }
-  return false;
+
+    return false;
 }
+

Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs

@@ -1,36 +1,51 @@
 /**
  * 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(n²) original, O(n) optimised using Set
+ * Space Complexity: O(n) for 2 options
+ * Optimal Time Complexity: O(n) linear time is the best possible
  *
  * @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;
+// }
+
 export function removeDuplicates(inputSequence) {
-  const uniqueItems = [];
+    const seenItems = new Set();
+    const uniqueSequence = [];
 
-  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;
-      }
+    for (const item of inputSequence) {
+        if (!seenItems.has(item)) {
+            seenItems.add(item);
+            uniqueSequence.push(item);
+        }
     }
-    if (!isDuplicate) {
-      uniqueItems.push(inputSequence[currentIndex]);
-    }
-  }
 
-  return uniqueItems;
+    return uniqueSequence;
 }
+

Sprint-1/Python/calculate_sum_and_product/calculate_sum_and_product.py

@@ -12,20 +12,32 @@ 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(2n) original (two passes), O(n) optimised (one pass)
+    Space Complexity: O(1) for both versions
+    Optimal time complexity: O(n) linear time is the best possible
     """
+    # 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:
+    #     product *= current_number
+
+    # return {"sum": sum, "product": product}
+
     # 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
+    total_sum = 0
+    total_product = 1
     for current_number in input_numbers:
-        product *= current_number
+        total_sum += current_number
+        total_product *= current_number
 
-    return {"sum": sum, "product": product}
+    return {"sum": total_sum, "product": total_product}