London | 25-SDC-NOV | Jesus del Moral | Sprint 1 | Analyse and Refactor Functions

Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js

@@ -9,21 +9,23 @@
  *   "product": 30 // 2 * 3 * 5
  * }
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity: First and Second loop both iterate through all n elements to calculate sum and product,
+ * O(n) + O(n) = O(2n) = O(n), Two sequential passes through the array
+ * Space Complexity: O(1), Only a constant amount of space is used for the sum and product variables, regardless of input size
+ * Optimal Time Complexity: O(n) - We must examine each element at least once to calculate the sum and product, so O(n) is the best we can achieve for this problem.
  *
+ * we can reduce the constant factor by eliminating the redundant second loop
+ * Instead of two passes (2n operations), we can do both calculations in a single pass (n operations)
+ * 
  * @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;
   }
 

Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs

@@ -1,34 +1,24 @@
 /**
  * Remove duplicate values from a sequence, preserving the order of the first occurrence of each value.
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * The complexity can be reduced from O(n^2) to O(n) by using a Set to track seen items,
+ * instead od searching through the entire UniqueItems array each time , we can check a Set in 0(1) time
+ * 
+ * Time Complexity: 0(n) Single pass through array and O(1) for Set lookups, resulting in O(n) overall
+ * Space Complexity: 0(n) Set and array both store up to n unique elements 
+ * Optimal Time Complexity: 0(n) - We must examine each element at least once to determine if it's a duplicate, so O(n) is the best we can achieve for this problem.
  *
  * @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/find_common_items/find_common_items.py

@@ -9,13 +9,11 @@ def find_common_items(
     """
     Find common items between two arrays.
 
-    Time Complexity:
-    Space Complexity:
-    Optimal time complexity:
+    Time Complexity: O(n + m) - Set creation and intersection 
+    Space Complexity:  O(n + m) - Two sets created for the input sequences
+    Optimal time complexity: O(n + m) - We must examine each element of both sequences at least once to find 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) & set(second_sequence))
+
+

Sprint-1/Python/has_pair_with_sum/has_pair_with_sum.py

@@ -7,12 +7,25 @@ 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: Single pass through the list of numbers, O(n), where n is the number of elements in the input list. Each lookup in the set is O(1) on average, so the overall time complexity is O(n).
+    Space Complexity: Set stores up to n elements in the worst case (if all numbers are unique), so the space complexity is O(n).
+    Optimal time complexity: 0(n) - We must examine each element at least once to determine if a pair exists that sums to the target, so O(n) is the best we can achieve for this problem.
     """
-    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
+
+    """seen is a set that will store the numbers we have already seen as we iterate through the list. 
+    For each number, we calculate its complement (the value needed to reach the target sum) 
+    and check if that complement is in the seen set. If it is, we have found a pair that sums to the target 
+    and can return True. If not, we add the current number to the seen set and continue iterating.
+     If we finish iterating through the list without finding a pair, we return False."""
+
+    seen = set() 
+
+    for num in numbers:
+        complement = target_sum - num
+
+        if complement in seen: 
+            return True
+
+        seen.add(num)
+
+    return False