London | 25-SDC-July | Andrei Filippov | 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) because this code has 2 loops that are separated from each other
+ * Space Complexity: O(n) for input array and O(1) for variables and loops, which result in O(n) in total
+ * Optimal Time Complexity: O(n) I think this algorithm is efficient enough
  *
  * @param {Array<number>} numbers - Numbers to process
  * @returns {Object} Object containing running total and product

Sprint-1/JavaScript/findCommonItems/findCommonItems.js

@@ -1,14 +1,29 @@
 /**
  * Finds common items between two arrays.
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity: O(n*m) or O(n^2) where n and m are length of each array
+ * Space Complexity: O(n) where n is a length of arrays
+ * Optimal Time Complexity: O(n)
  *
  * @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))),]
+  // program flow:
+  // firstArray.filter go through the each item in first array (loop)
+  // secondArray.includes check if that item is in the second array (nested loop)
+  // then, from new filtered array we create a Set which is not a nested array, but still O(n) operation
+  // nested loop result in O(n^2) time complexity
+
+  // to reduce time complexity to sublinear on the number of elements in the arrays we can use Set and it's .intersection() method
+
+  // complexity of this operation is O(n)
+  const firstArraySet = new Set(firstArray);
+  // complexity of this operation is O(m)
+  const secondArraySet = new Set(secondArray);
+  // intersection method uses has method which is considered to be faster than O(n) and since JS uses hash tables for has methods, the complexity should be
+  // has O(1) and intersection O(n) which result in O(n) overall time complexity
+  return Array.from(firstArraySet.intersection(secondArraySet));
+};

Sprint-1/Python/has_pair_with_sum/has_pair_with_sum.py

@@ -7,12 +7,30 @@ 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)
+    Space Complexity: O(n). It's memory for the input array arr
+    Optimal time complexity: 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
-    return False
+    # nested loop result in O(n^2) complexity
+    # 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
+
+    # it's better to use to pointers method here
+    # first: sort the list O(n)
+    numbers = sorted(numbers)
+    # declare the left and right pointers
+    left, right = 0, len(numbers)-1
+    # and until we "squeezed" the list check if sum of the pointers is equal to target sum
+    # in worst case we need to do n-1 steps where n is length of the list, so overall complexity will be O(n)
+    while left < right:
+        currentSum = numbers[left]+numbers[right]
+        if currentSum == target_sum:
+            return True
+        if currentSum > target_sum:
+            right -= 1
+        else:
+            left+=1
+    return False

Sprint-1/Python/remove_duplicates/remove_duplicates.py

@@ -7,19 +7,23 @@ 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)
+    Space complexity: O(n)
+    Optimal time complexity: 0(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
+    # nested array - O(n^2) complexity
+
+    # here we iterate through the array to make a dict which is O(n) and than convert it to the list which is also O(n)
+    return list(dict.fromkeys(values))