perf(js): optimize calculateSumAndProduct (2n → n)

Author: AliQassab

Sprint-1/JavaScript/findCommonItems/findCommonItems.js

@@ -1,14 +1,46 @@
 /**
  * Finds common items between two arrays.
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity: O(n + m) - Single pass through both arrays
+ * Space Complexity: O(min(n, m)) - Set size bounded by smaller array
+ * Optimal Time Complexity: O(n + m) - Cannot do better than linear time
  *
  * @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) => {
+  // OPTIMIZED IMPLEMENTATION: O(n + m) time complexity
+  // Previous implementation: O(n × m) due to nested includes() calls
+
+  // Convert second array to Set for O(1) lookup: O(m) time, O(m) space
+  const secondSet = new Set(secondArray);
+
+  // Find common items using Set lookup: O(n) time
+  const commonItems = firstArray.filter((item) => secondSet.has(item));
+
+  // Remove duplicates: O(n) time in worst case
+  return [...new Set(commonItems)];
+};
+
+/*
+ * ORIGINAL IMPLEMENTATION (for comparison):
+ *
+ * export const findCommonItems = (firstArray, secondArray) => [
+ *   ...new Set(firstArray.filter((item) => secondArray.includes(item))),
+ * ];
+ *
+ * COMPLEXITY ANALYSIS OF ORIGINAL:
+ * - firstArray.filter(): O(n) iterations
+ * - secondArray.includes(): O(m) for each iteration
+ * - Total: O(n × m) time complexity
+ * - Space: O(n) for Set creation
+ *
+ 
+ * IMPROVEMENTS MADE:
+ * 1. Reduced from O(n × m) to O(n + m) time complexity
+ * 2. Set lookup is O(1) vs Array.includes() O(m)
+ * 3. Significant performance gain for large arrays
+ * 4. Same functionality with better algorithmic efficiency
+ 
+ */

Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js

@@ -1,21 +1,59 @@
 /**
  * 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) - Single pass through the array
+ * Space Complexity: O(n) - Set to store seen numbers
+ * Optimal Time Complexity: O(n) - Cannot do better than linear time
  *
  * @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;
-      }
+  // OPTIMIZED IMPLEMENTATION: O(n) time complexity
+  // Previous implementation: O(n²) due to nested loops
+
+  const seen = new Set(); // O(n)
+
+  //  O(n) time complexity
+  for (const num of numbers) {
+    const complement = target - num;
+    // O(1) lookup
+    if (seen.has(complement)) {
+      return true;
     }
+
+    //  O(1) operation
+    seen.add(num);
   }
   return false;
 }
+console.log(hasPairWithSum([3, 2, 3, 4, 5], 9));
+/*
+ * ORIGINAL IMPLEMENTATION (for comparison):
+ *
+ * export function hasPairWithSum(numbers, target) {
+ *   for (let i = 0; i < numbers.length; i++) {           // O(n) iterations
+ *     for (let j = i + 1; j < numbers.length; j++) {     // O(n) iterations each
+ *       if (numbers[i] + numbers[j] === target) {       // O(1) comparison
+ *         return true;
+ *       }
+ *     }
+ *   }
+ *   return false;
+ * }
+ *
+ * COMPLEXITY ANALYSIS OF ORIGINAL:
+ * - Outer loop: O(n) iterations
+ * - Inner loop: O(n) iterations for each outer iteration
+ * - Total: O(n²) time complexity
+ * - Space: O(1) - only using loop variables
+ *
+ * PERFORMANCE ISSUES:
+ * - Quadratic time complexity O(n²)
+ *
+ * IMPROVEMENTS MADE:
+ * 1. Reduced from O(n²) to O(n) time complexity
+ * 2. Single pass through array instead of nested loops
+ * 3. Set lookup is O(1) vs nested iteration O(n)
+ */

Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs

@@ -1,36 +1,65 @@
 /**
  * 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) - Single pass through the array
+ * Space Complexity: O(n) - Set to track seen elements
+ * Optimal Time Complexity: O(n) - Cannot do better than linear time
  *
  * @param {Array} inputSequence - Sequence to remove duplicates from
  * @returns {Array} New sequence with duplicates removed
  */
 export function removeDuplicates(inputSequence) {
-  const uniqueItems = [];
+  // OPTIMIZED IMPLEMENTATION: O(n) time complexity
+  // Previous implementation: O(n²) due to nested loops checking each element
 
-  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]);
+  const seen = new Set(); // O(n)
+  const uniqueItems = []; // O(n)
+
+  //  O(n) time complexity
+  for (const item of inputSequence) {
+    //  O(1) lookup
+    if (!seen.has(item)) {
+      seen.add(item); // O(1) operation
+      uniqueItems.push(item); // O(1) operation
     }
   }
 
   return uniqueItems;
 }
+console.log(removeDuplicates([1, 2, 3, 4, 5, 1, 2, 3, 4, 5]));
+/*
+ * ORIGINAL IMPLEMENTATION (for comparison):
+ *
+ * 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;
+ * }
+ *
+ * COMPLEXITY ANALYSIS OF ORIGINAL:
+ * - Outer loop: O(n) iterations through input array
+ * - Inner loop: O(k) iterations through uniqueItems array (k grows with each unique element)
+ * - Worst case: O(n²) when all elements are unique
+ * - Space: O(n) for uniqueItems array
+ *
+ * PERFORMANCE ISSUES:
+ * - Quadratic time complexity O(n²) in worst case
+ *
+ * IMPROVEMENTS MADE:
+ * 1. Reduced from O(n²) to O(n) time complexity
+ * 2. Set lookup is O(1) vs linear search O(k)
+ * 3. Single pass through input array
+ */

Sprint-1/Python/calculate_sum_and_product/calculate_sum_and_product.py

@@ -12,20 +12,50 @@ 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) - Single pass through the list
+    Space Complexity: O(1) - Only using constant extra space
+    Optimal Time Complexity: O(n) - Cannot do better than linear time
     """
+    # OPTIMIZED IMPLEMENTATION: O(n) time complexity
+    # Previous implementation: O(2n) due to two separate loops
+    
+    sum_total = 0      # O(1) space
+    product = 1   # O(1) space 
+
+    #  O(n) time complexity
+    for current_number in input_numbers:
+        sum_total += current_number      # O(1) 
+        product *= current_number  # O(1) 
+
+    return {"sum": sum_total, "product": product}
+
+
+# ORIGINAL IMPLEMENTATION (for comparison):
+"""
+def calculate_sum_and_product(input_numbers: List[int]) -> Dict[str, int]:
     # Edge case: empty list
     if not input_numbers:
         return {"sum": 0, "product": 1}
 
     sum = 0
-    for current_number in input_numbers:
+    for current_number in input_numbers:    # First pass: O(n)
         sum += current_number
 
     product = 1
-    for current_number in input_numbers:
+    for current_number in input_numbers:     # Second pass: O(n)
         product *= current_number
 
-    return {"sum": sum, "product": product}
+    return {"sum": sum, "product": product}  # Total: O(2n) = O(n) time
+
+COMPLEXITY ANALYSIS OF ORIGINAL:
+- First loop: O(n) iterations to calculate sum
+- Second loop: O(n) iterations to calculate product
+- Total: O(2n) = O(n) time complexity
+- Space: O(1) - only using loop variables
+
+IMPROVEMENTS MADE:
+1. Reduced from 2n to n operations (50% fewer iterations)
+2. Single pass through list instead of two separate loops
+3. Same O(n) time complexity but with better constant factors
+"""

Sprint-1/Python/find_common_items/find_common_items.py

@@ -9,13 +9,49 @@ def find_common_items(
     """
     Find common items between two arrays.
 
-    Time Complexity:
-    Space Complexity:
-    Optimal time complexity:
+    Time Complexity: O(n + m) - Single pass through both sequences
+    Space Complexity: O(min(n, m)) - Set size bounded by smaller sequence
+    Optimal Time Complexity: O(n + m) - Cannot do better than linear time
     """
+    # OPTIMIZED IMPLEMENTATION: O(n + m) time complexity
+    # Previous implementation: O(n × m) due to nested loops with linear search
+    
+    # Convert second sequence to set for O(1) lookup: O(m) time, O(m) space
+    second_set = set(second_sequence)
+    
+    # Find common items using set lookup: O(n) time
+    common_items = []
+    for item in first_sequence:
+        if item in second_set and item not in common_items:
+            common_items.append(item)
+    
+    return common_items
+
+
+# ORIGINAL IMPLEMENTATION (for comparison):
+"""
+def find_common_items(
+    first_sequence: Sequence[ItemType], second_sequence: Sequence[ItemType]
+) -> List[ItemType]:
     common_items: List[ItemType] = []
-    for i in first_sequence:
-        for j in second_sequence:
-            if i == j and i not in common_items:
+    for i in first_sequence:                    # O(n) iterations
+        for j in second_sequence:               # O(m) iterations each
+            if i == j and i not in common_items: # O(k) linear search
                 common_items.append(i)
     return common_items
+
+COMPLEXITY ANALYSIS OF ORIGINAL:
+- Outer loop: O(n) iterations through first_sequence
+- Inner loop: O(m) iterations through second_sequence
+- Linear search: O(k) for 'i not in common_items' check
+- Total: O(n × m × k) time complexity in worst case
+- Space: O(n) for common_items list
+
+PERFORMANCE ISSUES:
+- Quadratic time complexity O(n × m) from nested loops
+
+IMPROVEMENTS MADE:
+1. Reduced from O(n × m × k) to O(n + m) time complexity
+2. Set lookup is O(1) vs nested iteration O(m)
+3. Single pass through first_sequence
+"""

Sprint-1/Python/has_pair_with_sum/has_pair_with_sum.py

@@ -7,12 +7,48 @@ 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) - Single pass through the list
+    Space Complexity: O(n) - Set to store seen numbers
+    Optimal Time Complexity: O(n) - Cannot do better than linear time
     """
-    for i in range(len(numbers)):
-        for j in range(i + 1, len(numbers)):
-            if numbers[i] + numbers[j] == target_sum:
+    # OPTIMIZED IMPLEMENTATION: O(n) time complexity
+    # Previous implementation: O(n²) due to nested loops
+    
+    seen = set()  # O(n) space for storing seen numbers
+    
+    # Single pass through list: O(n) time complexity
+    for num in numbers:
+        complement = target_sum - num  
+        
+        # Check if complement exists in seen numbers: O(1) lookup
+        if complement in seen:
+            return True  
+        
+        # Add current number to seen set: O(1) operation
+        seen.add(num)
+    
+    return False  
+
+# ORIGINAL IMPLEMENTATION (for comparison):
+"""
+def has_pair_with_sum(numbers: List[Number], target_sum: Number) -> bool:
+    for i in range(len(numbers)):           # O(n) iterations
+        for j in range(i + 1, len(numbers)): # O(n) iterations each
+            if numbers[i] + numbers[j] == target_sum:  # O(1) comparison
                 return True
     return False
+
+COMPLEXITY ANALYSIS OF ORIGINAL:
+- Outer loop: O(n) iterations
+- Inner loop: O(n) iterations for each outer iteration
+- Total: O(n²) time complexity
+- Space: O(1) - only using loop variables
+
+PERFORMANCE ISSUES:
+- Quadratic time complexity O(n²)
+
+IMPROVEMENTS MADE:
+1. Reduced from O(n²) to O(n) time complexity
+2. Single pass through list instead of nested loops
+3. Set lookup is O(1) vs nested iteration O(n)
+"""