London | 26-SDC-July | Raihan Sharif | Sprint 1 | Analyse and Refactor Functions

Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js

@@ -9,26 +9,23 @@
  *   "product": 30 // 2 * 3 * 5
  * }
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity: O(n) -- Two loops through array.
+ * Space Complexity: 0(1) -- sum, product, num are all fixed size.
+ * Optimal Time Complexity: O(n) -- every element can be read once, but in single loop.
  *
  * @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 sum = 0;
+    let product = 1;
+    for (const num of numbers) {
+        sum += num;
+        product *= num;
+    }
 
-  let product = 1;
-  for (const num of numbers) {
-    product *= num;
-  }
-
-  return {
-    sum: sum,
-    product: product,
-  };
+    return {
+        sum: sum,
+        product: 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) -- where n is length of firstArray and
+ * m is length of secondArray. filter is O(n) and includes is o(m) for every n.
+ *
+ * Space Complexity: O(n) -- every element in firstArray could have its equivalent in secondArray.
+ * So in the worst case, n elements would have to be stored in the filtered array.
+ * Size of filtered array is k, such that k <= n. the Set (l) and the final spread array (m) are
+ * also smaller than n. The space used is k + l + m, so ultimately it's still O(n).
+ *
+ * Optimal Time Complexity: O(n + m) -- converting secondArray to a set gives you a O(1) lookup,
+ * with secondSet.has(item). So filter becomes O(n * 1) rather than O(n * m).
+ *
+ * Converting an array to a Set is a O(m) operation, since every element in secondArray (size of m),
+ * must be checked. As this only done once at the start, rather than for every element in firstArray.
+ * The final time complexity is O(n + m).
+ *
+ * However, there is increased space complexity of O(n + m) as a set of size m now needs to be stored.
  *
  * @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) => {
+    // dedupe the secondArray to reduce the number of inclues()
+    const secondSet = new Set(secondArray);
+    return [...new Set(firstArray.filter((item) => secondSet.has(item)))];
+};

Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js

@@ -1,21 +1,28 @@
 /**
  * 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) -- if there is no pair that their sum = target, 
+ * or its is the last two items then the function has to loop over every element, 
+ * and for each element loop n-1, n-2, n-3, ..., 0 times.
+ * Total comparisons is n(n-1)/2 = (n^2 - n)/2. In big O terms that's just O(n^2).
+ * 
+ * Space Complexity: O(1) -- space used does not grow with input size.
+ * 
+ * Optimal Time Complexity: O(n) -- At worst, makes one pass through the array.
+ * All operations inside of the loop are O(1). 
  *
  * @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;
-      }
+  const seen = new Set();
+
+  for (const num of numbers) {
+    if (seen.has(target - num)) {
+      return true;
     }
+    seen.add(num);
   }
   return false;
 }

Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs

@@ -1,36 +1,20 @@
 /**
  * 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) -- In the worst case, the number of comparisons in each step of the iteration
+ * increases by 1 because uniqueItems grows by 1. It goes 0 + 1 + 2 + ... + (n-1), so total number of  
+ * comparisons is n(n - 1)/2 = (n^2 - n)/2. Big O discards the constant factor (1/2) and the lower-order term (n), leaving n^2.
+ * 
+ * There aren't exactly n^2 comparisons, but big O doesn't care about that.
+ * 
+ * Space Complexity: O(n) -- uniqueItems can hold up to n items when all items are unique.
+ * 
+ * Optimal Time Complexity: O(n) -- Adding each element of the input sequence to the set is a O(1) (on average) operation,
+ * and then another O(n) operation to convert to an array with the spread operation.
  *
  * @param {Array} inputSequence - Sequence to remove duplicates from
  * @returns {Array} New sequence with duplicates removed
  */
 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]);
-    }
-  }
-
-  return uniqueItems;
+  return [...new Set(inputSequence)];
 }

Sprint-1/Python/calculate_sum_and_product/calculate_sum_and_product.py

@@ -12,20 +12,24 @@ 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) -- Two separate loops through the list so 2n, but big O does not care
+    about the coefficient, so the complexity is O(n).
+
+    Space Complexity: O(1) -- Three fixed size variables (sum, product, and current_number). The output
+    dict is also fixed size regardless of input size. 
+
+    Optimal time complexity: O(n) -- Use one loop instead of two, cuts the time taken by half, but the relation
+    between input size and computation time remains the same.
     """
     # 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 +9,25 @@ def find_common_items(
     """
     Find common items between two arrays.
 
-    Time Complexity:
-    Space Complexity:
-    Optimal time complexity:
+    Time Complexity: O(n * m * k) -- where n = length of first_sequence, m = length of second_sequence,
+    k = length of common_items. k <= min(n, m) because common_items can never be bigger than the smaller 
+    of the two sequences. As common_items is a list, checking membership is a linear time operation. 
+
+    Complexity class is O(n^2) if two are sequences are same size.
+
+    Space Complexity: O(n) -- If every item in first_sequence is also in second_sequence, then the entirety of n 
+    has to be stored in common_items. Techincally it's O(min(n, m)).
+
+    Optimal time complexity: O(m + n) -- O(m) to convert second_sequence to a set, then. O(n) to iterate through 
+    the first_sequence. All operations inside the loop are O(1) (more or less), so additional complexity.
     """
+    second_set = set(second_sequence)   # O(m)
+    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)
+
+    for item in first_sequence:                       # n iterations
+        if item in second_set and item not in seen:   # O(1) average lookups, times 2
+            seen.add(item)                            # O(1) average insertion              
+            common_items.append(item)                 # O(1)
+
     return common_items

Sprint-1/Python/has_pair_with_sum/has_pair_with_sum.py

@@ -7,12 +7,20 @@ 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) -- if there is no pair that their sum = target_sum, 
+    or its is the last two items then the function has to loop over every element, 
+    and for each element loop n-1, n-2, n-3, ..., 0 times.
+    Total comparisons is n(n-1)/2 = (n^2 - n)/2. In big O terms that's just O(n^2).
+
+    Space Complexity: O(1) -- no additional memory is used.
+
+    Optimal time complexity: O(n) -- Goes through the list at most once. adding to set is 
+    O(1). 
     """
-    for i in range(len(numbers)):
-        for j in range(i + 1, len(numbers)):
-            if numbers[i] + numbers[j] == target_sum:
-                return True
+    seen = set()
+    for num in numbers:
+        # check if there is a number in the set that added to num = target_sum
+        if target_sum - num in seen: 
+            return True
+        seen.add(num)
     return False

Sprint-1/Python/remove_duplicates/remove_duplicates.py

@@ -7,19 +7,16 @@ 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:
-    """
-    unique_items = []
+    Time complexity: O(n^2) -- In the worst case, when all items are unique, the number of comparisons 
+    in each step of the iteration increases by 1 because unique_items grows by 1. 
+    It goes 0 + 1 + 2 + ... + (n-1), so total number of comparisons is n(n - 1)/2 = (n^2 - n)/2. 
+    Big O discards the constant factor (1/2) and the lower-order term (n), leaving n^2. 
 
-    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)
+    Space complexity: O(n) -- unique_items can hold up to n items when all items are unique.
+
+    Optimal time complexity: O(n) -- Adding each element of "values" to the dict is a O(1) (on average) operation,
+    and then another O(n) operation to convert to a list.
+    """
 
-    return unique_items
+    # creates a dictionary, using the values of the input as as keys (unique, and preserves order)
+    return list(dict.fromkeys(values))