NW | 26-SDC-Mar | TzeMing Ho | Sprint 1 | Analyse and Refactor Functions

Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js

@@ -9,9 +9,9 @@
  *   "product": 30 // 2 * 3 * 5
  * }
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity: O(N), as the function iterates the array twice (2N)
+ * Space Complexity: O(1), the space for variables remains constant
+ * Optimal Time Complexity: O(N), the function can't be refactored to reduce complexity, (N) is regarded as the same as (2N)
  *
  * @param {Array<number>} numbers - Numbers to process
  * @returns {Object} Object containing running total and product

Sprint-1/JavaScript/findCommonItems/findCommonItems.js

@@ -1,14 +1,22 @@
 /**
  * Finds common items between two arrays.
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity: O(N*M), for each item of loop in the firstArray, it has to loop through the secondArray in the worst case
+ * Space Complexity: O(N), the worst case is the length of the firstArray
+ * Optimal Time Complexity: O(N + M), N is for the Set of firstArray, and M is the loop for secondArray. O(N+M) is better than O(N*M). 
  *
  * @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))),
+// ];
+
+export const findCommonItems = (firstArray, secondArray) => {
+  const firstSet = new Set(firstArray);
+
+  const commonInSecond = secondArray.filter((item) => firstSet.has(item));
+
+  return [... new Set(commonInSecond)];
+}

Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js

@@ -1,21 +1,34 @@
 /**
  * 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^2), the loop is inside a loop
+ * Space Complexity: O(1). O(i) is for i and j index counters.
+ * Optimal Time Complexity: O(N), it loops once after optimization. However, the space complexity becomes O(N) as there is a growing targetSet.
  *
  * @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) {
-  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 targetSet = new Set();
+  for (let num of numbers) {
+    const targetNum = target - num;
+    if (targetSet.has(targetNum)) {
+      return true;
+    } else {
+      targetSet.add(num)
     }
   }
   return false;
-}
+}

Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs

@@ -1,36 +1,40 @@
 /**
  * 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^2). It takes a loop inside a loop to compare duplicates.
+ * Space Complexity: O(N). The uniqueItems array would grow to N in the worst case.
+ * Optimal Time Complexity: O(N). It only loops once after optimization. 
  *
  * @param {Array} inputSequence - Sequence to remove duplicates from
  * @returns {Array} New sequence with duplicates removed
  */
-export function removeDuplicates(inputSequence) {
-  const uniqueItems = [];
+// 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]);
+//     }
+//   }
 
-  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;
+// }
 
-  return uniqueItems;
-}
+export function removeDuplicates(inputSequence) {
+  return [...new Set(inputSequence)];
+}

Sprint-1/Python/calculate_sum_and_product/calculate_sum_and_product.py

@@ -12,9 +12,9 @@ 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 function loops the array for twice (2N)
+    Space Complexity: O(1), the space for variables remains constant
+    Optimal time complexity: O(N), (N) is regarded as the same as (2N)
     """
     # Edge case: empty list
     if not input_numbers:

Sprint-1/Python/find_common_items/find_common_items.py

@@ -9,13 +9,16 @@ def find_common_items(
     """
     Find common items between two arrays.
 
-    Time Complexity:
-    Space Complexity:
-    Optimal time complexity:
+    Time Complexity: O(N*M), it is a loop inside a loop
+    Space Complexity: O(N), the size of the common_items would grow to N in the worst case
+    Optimal time complexity: O(N + M), N is for the firstSet, M is the filter in second
     """
-    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
+    # 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
+
+    return list(set(first_sequence).intersection(second_sequence))
+

Sprint-1/Python/has_pair_with_sum/has_pair_with_sum.py

@@ -7,12 +7,22 @@ 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), a loop inside a loop
+    Space Complexity: O(1). O(1) is for the i and j loop counter.
+    Optimal time complexity: O(N) as it only loops once, with a trade off space becomes 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
+    # 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
+
+    pair_num_set = set()
+
+    for num in numbers:
+        pair_num = target_sum - num
+        if pair_num in pair_num_set:
+            return True
+        else:
+            pair_num_set.add(num)
     return False

Sprint-1/Python/remove_duplicates/remove_duplicates.py

@@ -7,19 +7,21 @@ 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^2). It takes a loop inside a loop
+    Space complexity: O(N). The space of unique_items would grow to N in the worst case.
+    Optimal time complexity: O(N). It only loops once after optimization, but the space becomes O(N + N)
     """
-    unique_items = []
+    # 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
+
+    return list(dict.fromkeys(values))