London | 26-SDC-Mar | Ebrahim Beiati-Asl | Sprint 1 | Sprint 1 analyse and refactor functions

Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js

@@ -9,26 +9,29 @@
  *   "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)
+ * Explanation: The function iterates through the list of numbers twice,
+ *  once to calculate the sum and once to calculate the product. 
+ * Each iteration takes O(n) time, resulting in a total time complexity of O(n). 
+ * The space complexity is O(1) because we are using a constant amount of space to store the sum and product, 
+ * regardless of the size of the input list.
+ * 
+ * Refactor:
+ * I combined the two loops into a single loop 
+ * to calculate both the sum and product simultaneously.
  *
  * @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;
   }
 
-  return {
-    sum: sum,
-    product: product,
-  };
+  return {sum,product};
 }

Sprint-1/JavaScript/findCommonItems/findCommonItems.js

@@ -1,14 +1,26 @@
 /**
  * Finds common items between two arrays.
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
- *
+ * Time Complexity: O(n * m)
+ * Space Complexity: O(k)---> where k is the number of common items
+ * Optimal Time Complexity: O(n + m)
+ * Explanation: The function uses the filter method to iterate through each item in the first array and
+ * checks if it exists in the second array using the includes method.
+ * This results in a time complexity of O(n * m) because for each item in the first array,
+ * we are checking against all items in the second array.
+ *  The space complexity is O(k) because we are creating a new array to store the common items,
+ *  where k is the number of common items found.
  * @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 lookup = new Set(secondArray);
+  const result = new Set();
+  for (const item of firstArray) {
+    if (lookup.has(item)) {
+      result.add(item);
+    }
+  }
+  return [...result];
+};

Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js

@@ -1,21 +1,26 @@
 /**
  * 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)
+ * Space Complexity: O(1)
+ * Optimal Time Complexity: O(n)
+ * Explanation: The function uses a nested loop to check every possible pair of numbers in the array
+ *  to see if they sum up to the target value.
+ * This results in a time complexity of O(n²) because for each number in the array,
+ * we are checking it against every other number.
+ *The refactored version uses a Set to check for the existence of the complement, reducing the overall complexity to 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;
-      }
+  const seen = new Set();
+  for (const num of numbers) {
+    const complement = target - num;
+    if (seen.has(complement)) {
+      return true;
     }
+    seen.add(num);
   }
   return false;
 }

Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs

@@ -1,36 +1,27 @@
 /**
  * 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)
+ * Space Complexity: O(n)
+ * Optimal Time Complexity: O(n)
+ * Explanation: The function checks each element against all previously seen elements to determine,
+ * if it is a duplicate, resulting in a time complexity of O(n²) in the worst case.
+ * The space complexity is O(n) because we are storing unique items in a new array.
+ *  The optimal time complexity can be achieved by using a Set to track seen items,
+ * which allows for O(1) average time complexity for lookups, resulting in an overall time complexity of O(n).
+ * Refactor: Using Set allows to constant-time membership checks, which significantly reduces the time complexity from O(n²) to O(n).
  *
  * @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);
     }
   }
-
   return uniqueItems;
 }

Sprint-1/Python/calculate_sum_and_product/calculate_sum_and_product.py

@@ -12,20 +12,24 @@ 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)
+    Explanation: The function loops through the list of input numbers twice, once to calculate the
+    sum and once to calculate the product. The results in O(n n) = O(n) time. It uses only
+
+    Refactor: We can calculate both the sum and product in a single loop,
+    which reduces the time complexity to 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
+    total_sum = 0
+    total_product = 1
 
-    product = 1
-    for current_number in input_numbers:
-        product *= current_number
+    for num in input_numbers:
+        total_sum += num
+        total_product *= num
 
-    return {"sum": sum, "product": product}
+    return {"sum": total_sum, "product": total_product}

Sprint-1/Python/find_common_items/find_common_items.py

@@ -9,13 +9,22 @@ def find_common_items(
     """
     Find common items between two arrays.
 
-    Time Complexity:
-    Space Complexity:
-    Optimal time complexity:
+    Time Complexity: O(n * m)
+    Space Complexity: O(n)
+    Optimal time complexity: O(n + m)
+    Explanation: The function uses a nested loop to check every possible pair of items in the two
+    sequences to see if they are common.
+    This results in a time complexity of O(n * m).
+    Using a set for constant-time lookups can reduce the time complexity to O(n + m)
+    by first converting one of the sequences into a set and then checking for common items in a single loop.
     """
-    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
+    lookup = set(second_sequence)
+    result = []
+    seen = set()
+
+    for item in first_sequence:
+        if item in lookup and item not in seen:
+            seen.add(item)
+            result.append(item)
+
+    return result

Sprint-1/Python/has_pair_with_sum/has_pair_with_sum.py

@@ -7,12 +7,17 @@ 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^2) because we have two nested loops that iterate through the list of numbers.
+    Space Complexity: O(1)
+    Optimal time complexity: O(n)
+    Explanation: The function checks every possible pair of numbers using two nested loops, resulting in O(n^2) time complexity.
+    Refactor: It uses constant extra space. The optimal solution uses a set to check complements in O(1) time, reducing the overall complexity to 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:
+        complement = target_sum - num
+        if complement in seen:
+            return True
+        seen.add(num)
+
     return False

Sprint-1/Python/remove_duplicates/remove_duplicates.py

@@ -6,20 +6,18 @@
 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:
+    Outer loop runs n times (each value in values)
+    Inner loop runs up to k times (size of unique_items, worst case k = n)
+    Time complexity: O(n^2)
+    Space complexity: O(n)
+    Optimal time complexity: O(n)
     """
-    unique_items = []
+    seen = set()
+    result = []
 
     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)
+        if value not in seen:
+            seen.add(value)
+            result.append(value)
 
-    return unique_items
+    return result