London | Ameneh Keshavarz | Module-Complexity | Sprint 1

Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js

@@ -9,14 +9,13 @@
  *   "product": 30 // 2 * 3 * 5
  * }
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity: O(n)
+ * Space Complexity: O(1) because we are only using two variables(sum,product) and the space does not grow with input
  *
  * @param {Array<number>} numbers - Numbers to process
  * @returns {Object} Object containing running total and product
  */
-export function calculateSumAndProduct(numbers) {
+/*export function calculateSumAndProduct(numbers) {
   let sum = 0;
   for (const num of numbers) {
     sum += num;
@@ -27,8 +26,26 @@ export function calculateSumAndProduct(numbers) {
     product *= num;
   }
 
+  return {
+    sum: sum,
+    product: product,
+  };
+}*/
+
+//Optimal Time Complexity: O(n)
+export function calculateSumAndProduct(numbers) {
+  let sum = 0;
+  let product = 1;
+
+  for (const num of numbers) { //O(n)
+    sum += num;
+    product *= num;
+  }
+
   return {
     sum: sum,
     product: product,
   };
 }
+
+// this solution iterates once instead of two 

Sprint-1/JavaScript/findCommonItems/findCommonItems.js

@@ -1,14 +1,34 @@
 /**
  * Finds common items between two arrays.
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity:O(n * m) because for each element in firstArray, the includes method does a linear search in the second array
+ *  which can take up to m steps, so that would be O(n*m)
+ * 
+ * Space Complexity: O(n) beacause firstArray.filter() creates an array of at most n items and Set stores at most n items, 
+ * so the overall complexity would be O(n)+O(n) which we consider O(n) as they grow linerly not exponentially
+ * 
  *
  * @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) => [
+/*export const findCommonItems = (firstArray, secondArray) => [
   ...new Set(firstArray.filter((item) => secondArray.includes(item))),
-];
+];*/
+
+//Optimal Time Complexity: O(n+m+k)
+
+export const findCommonItems = (firstArray, secondArray) => {
+  const setSecond = new Set(secondArray); // O(m)
+  const seen = new Set();
+
+  for (const item of firstArray) { // )(n)
+    if (setSecond.has(item)) {
+      seen.add(item);
+    }
+  }
+
+  return [...seen]; //)(k) this creates an array from seen which contains only the common items
+};
+
+// this solution is more optimised as it reduces Time Complexity. The time complexity of the inital function was quadratic O(n*m) but now it's linear O(n+m+k)

Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js

@@ -1,21 +1,37 @@
 /**
  * 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(n²) there are two nested for loops 
+ * Space Complexity: O(1) becuae we are not using any extra space
  *
  * @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) {
+/*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;
       }
     }
   }
+  return false;
+}*/
+
+//Optimal Time Complexity: O(n)
+
+export function hasPairWithSum(numbers, target) {
+  const seen = new Set();
+
+  for (const num of numbers) { // O(n)
+    const complement = target - num;
+    if (seen.has(complement)) {
+      return true;
+    }
+    seen.add(num);
+  }
+
   return false;
 }
+// the time complexity of original solution was quadratic O(n²) but it's linear for the second solution O(n)

Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs

@@ -1,14 +1,13 @@
 /**
  * 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(n²) because we have two nested loops
+ * Space Complexity: O(n) uniqueItems array stores up to n elements if all are unique
  *
  * @param {Array} inputSequence - Sequence to remove duplicates from
  * @returns {Array} New sequence with duplicates removed
  */
-export function removeDuplicates(inputSequence) {
+/*export function removeDuplicates(inputSequence) {
   const uniqueItems = [];
 
   for (
@@ -32,5 +31,23 @@ export function removeDuplicates(inputSequence) {
     }
   }
 
+  return uniqueItems;
+}*/
+
+//Optimal Time Complexity: O(n) we iterate through the inputSequence once
+
+export function removeDuplicates(inputSequence) {
+  const seen = new Set();
+  const uniqueItems = [];
+
+  for (const item of inputSequence) {
+    if (!seen.has(item)) {
+      seen.add(item);
+      uniqueItems.push(item);
+    }
+  }
+
   return uniqueItems;
 }
+// the time complexity of original solution was quadratic O(n²) but the it's linear for the second solution
+