diff --git a/Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js b/Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js
index ce738c3..1691fc4 100644
--- a/Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js
+++ b/Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js
@@ -9,22 +9,19 @@
  *   "product": 30 // 2 * 3 * 5
  * }
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity: is o(n) - coz only we have one input that needs at least one loop 
+ * Space Complexity: here the space is o(1) coz the memory will not growing whatever happen with input 
+ * Optimal Time Complexity: here even when i drop one of the loops the optimal time complexity will be still O(n) because we dealing with input need at least one 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 product = 1;
   for (const num of numbers) {
-    product *= num;
+    sum += num;
+    product *= num
   }
 
   return {
@@ -32,3 +29,6 @@ export function calculateSumAndProduct(numbers) {
     product: product,
   };
 }
+
+
+console.log(calculateSumAndProduct([2, 3, 8, 9]));
diff --git a/Sprint-1/JavaScript/findCommonItems/findCommonItems.js b/Sprint-1/JavaScript/findCommonItems/findCommonItems.js
index 5619ae5..11fbaf6 100644
--- a/Sprint-1/JavaScript/findCommonItems/findCommonItems.js
+++ b/Sprint-1/JavaScript/findCommonItems/findCommonItems.js
@@ -1,14 +1,17 @@
 /**
  * Finds common items between two arrays.
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity: now is O(n^2) - coz we doing instead loop one with the filter first array then we checking the array with includes loop to find out if the item are common in both arrays 
+ * Space Complexity: is O(n) - coz it is grows directly proportional with the input size 
+ * Optimal Time Complexity: to O(n), after i used a Set, now the lookup is faster instead of checking the whole second array each 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) => {
+  let second = new Set(secondArray)
+  return [ ...new Set(firstArray.filter((item) => second.has(item))) ]
+}
+console.log(findCommonItems([1, 2, 2, 3], [4, 3, 2]))
+
diff --git a/Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js b/Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js
index dd2901f..d82c307 100644
--- a/Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js
+++ b/Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js
@@ -1,21 +1,27 @@
 /**
  * Find if there is a pair of numbers that sum to a given target value.
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity: is O(n^2) coz we have nested loop in the function
+ * Space Complexity: is O(n) coz it grows proportionally with the input size    
+ * Optimal Time Complexity: to O(n) - and that by using Set as check point to fast lookup and catch up if it has the number instead of re-scanning everything
  *
  * @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 checkPoint = new Set()
+
+  for(let i=0; i < numbers.length; i++){
+    let num = target - numbers[i]
+    if(checkPoint.has(num)){
+      return true
+    }else {
+      checkPoint.add(numbers[i])
     }
   }
+  
   return false;
 }
+
+console.log(hasPairWithSum([1, 2, 4, 5, 3], 9));
diff --git a/Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs b/Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs
index dc5f771..d306749 100644
--- a/Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs
+++ b/Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs
@@ -1,36 +1,18 @@
 /**
  * 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: is O(n^2) coz it used nested loop to remove the duplications 
+ * Space Complexity: it is O(n) coz it is grows directly proportional with the input size 
+ * Optimal Time Complexity: to O(n) - because even if i'm using Set, i need to go through each element in the array, which leads me to the time complexity will be growing directly proportional with the input size
  *
  * @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]);
-    }
-  }
+  const uniqueItems = [...new Set(inputSequence)]
 
   return uniqueItems;
 }
+
+console.log(removeDuplicates([5, 2, 2, 3, 4, 4, 1]))
+console.log(removeDuplicates(["a", "e", "b", "a", "c", "c", "a"]))