streamline calculate_sum_and_product to use a single loop for efficiency

Author: Ify-O

Sprint-1/Python/calculate_sum_and_product/calculate_sum_and_product.py

@@ -5,27 +5,32 @@ def calculate_sum_and_product(input_numbers: List[int]) -> Dict[str, int]:
     """
     Calculate the sum and product of integers in a list.
 
-    Note: the sum is every number added together
-    and the product is every number multiplied together
-    so for example: [2, 3, 5] would return
-    {
-        "sum": 10, // 2 + 3 + 5
-        "product": 30 // 2 * 3 * 5
-    }
-    Time Complexity:
-    Space Complexity:
-    Optimal time complexity:
-    """
-    # Edge case: empty list
-    if not input_numbers:
-        return {"sum": 0, "product": 1}
+    Time Complexity: O(n)
+    Space Complexity: O(1)
+    Optimal Time Complexity: O(n)
+
+    Refactor:
+    The original implementation iterated through the list twice:
+    once to calculate the sum and once to calculate the product.
+    This refactor combines both calculations into a single loop,
+    reducing the number of iterations while maintaining O(n)
+    time complexity.
 
-    sum = 0
-    for current_number in input_numbers:
-        sum += current_number
+    Args:
+        input_numbers: List of integers.
 
+    Returns:
+        Dictionary containing the sum and product.
+    """
+
+    total_sum = 0
     product = 1
-    for current_number in input_numbers:
-        product *= current_number
 
-    return {"sum": sum, "product": product}
+    for number in input_numbers:
+        total_sum += number
+        product *= number
+
+    return {
+        "sum": total_sum,
+        "product": product,
+    }