London | 25-SDC-Nov | Emiliano Uruena | Sprint 1 | Analyse and Refactor Functions

Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs

@@ -1,36 +1,19 @@
 /**
  * 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: for each element in the input is compared against all previous unique elements: o(n2) 
+ * Space Complexity: we store n elements it all elements are unique.
+ * Optimal Time Complexity:  O(n) — You must inspect each element at least once, and Set lookups are O(1).
  *
  * @param {Array} inputSequence - Sequence to remove duplicates from
  * @returns {Array} New sequence with duplicates removed
  */
 export function removeDuplicates(inputSequence) {
-  const uniqueItems = [];
+  const seenItems = new Set();
 
-  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 value of inputSequence) {
+      seenItems.add(value);
   }
 
-  return uniqueItems;
+  return Array.from(seenItems);
 }

Sprint-1/Python/calculate_sum_and_product/calculate_sum_and_product.py

@@ -13,19 +13,21 @@ def calculate_sum_and_product(input_numbers: List[int]) -> Dict[str, int]:
         "product": 30 // 2 * 3 * 5
     }
     Time Complexity:
+     There were two loops, one for each operation. Each has complexity O(n)
+     for both loops, complexity is O(2n)
     Space Complexity:
+        space is constant O(1)
     Optimal time complexity:
+        With one loop for both operations, must iterate through all n elements in the list, so the overall time complexity is 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
-
     product = 1
     for current_number in input_numbers:
+        sum += current_number
         product *= current_number
 
     return {"sum": sum, "product": product}

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: The time complexity is O(n + m) because we build a set from the second sequence in O(m) time and iterate through the first sequence in O(n) time.
+    Space Complexity: The space complexity is O(n + m) because we store up to all elements of the second sequence in a set and up to the common elements in additional storage. 
+    Optimal time complexity: The time complexity is O(n + m) and the space complexity is O(n + m).
     """
+    second_set = set(second_sequence)
+    seen = set()
     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)
+        if i in second_set and i not in seen:
+            seen.add(i)
+            common_items.append(i)
+
     return common_items

Sprint-1/Python/has_pair_with_sum/has_pair_with_sum.py

@@ -8,11 +8,15 @@ 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:
+    there are loops nested that check every pair O(n2) 
+    Space Complexity: O(n)
     Optimal time complexity:
+     using a set to check numbers seen 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
+    numbers_seen = set()
+    for num in numbers:  # O(n)
+        required_num = target_sum - num
+        if required_num in numbers_seen:
+            return True
+        numbers_seen.add(num)
     return False