diff --git a/Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js b/Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js
index ce738c33..1e070ca6 100644
--- a/Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js
+++ b/Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js
@@ -9,21 +9,22 @@
  *   "product": 30 // 2 * 3 * 5
  * }
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Time Complexity: O(n)
+ * Space Complexity:O(1)
+ * Optimal Time Complexity:O(n)
  *
  * @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;
+
+  // Initial code looped twice through the array, i.e. 2 operations (n + n) = O(2n)operations => O(n). Limiting the loop to just one operation makes it n operation => O(n).
+  // The space complexity is constant as the variables do not grow when the function is called.
+
   for (const num of numbers) {
+    sum += num;
     product *= num;
   }
 
diff --git a/Sprint-1/JavaScript/findCommonItems/findCommonItems.js b/Sprint-1/JavaScript/findCommonItems/findCommonItems.js
index 5619ae5d..6dc77e25 100644
--- a/Sprint-1/JavaScript/findCommonItems/findCommonItems.js
+++ b/Sprint-1/JavaScript/findCommonItems/findCommonItems.js
@@ -1,14 +1,21 @@
 /**
  * Finds common items between two arrays.
  *
- * Time Complexity:
- * Space Complexity:
- * Optimal Time Complexity:
+ * Initial code has this and can be refactored for optimisation
+ * Time Complexity: O(nm)
+ * Space Complexity: O(n)
+ * Optimal Time Complexity: O(n + m)
  *
  * @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))),
+// ];
+// The initial solution had an expensive operation of .includes(item) search on the secondArray, resulting in O(nm) time complexity. The optimised code stores the secondArray in a set and allows O(1) lookup and reduce the complexity to O(n+m)
+export const findCommonItems = (firstArray, secondArray) => {
+  const secondSet = new Set(secondArray);
+
+  return [...new Set(firstArray.filter((item) => secondSet.has(item)))];
+};
diff --git a/Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js b/Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js
index dd2901f6..97bac660 100644
--- a/Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js
+++ b/Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js
@@ -1,21 +1,24 @@
 /**
  * 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)
+ * Space Complexity:O(1)
+ * Optimal Time Complexity: O(n)
  *
  * @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) {
+  const seenNumbers = new Set();
   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;
-      }
+    // initial solution iterates twice, having outer and inner loop which gives us a time complexity of O(n**2). To optimise this, using a set to store seen numbers, then check whether the complement(target - number) for each num in the array has been seen and in the set. This gives a complexity of O(n)
+    let complement = target - numbers[i];
+
+    if (seenNumbers.has(complement)) {
+      return true;
     }
+    seenNumbers.add(numbers[i]);
   }
   return false;
 }
diff --git a/Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs b/Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs
index dc5f7711..cfb1215f 100644
--- a/Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs
+++ b/Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs
@@ -1,36 +1,22 @@
 /**
  * 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: O(n)
  *
  * @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 seenItems = new Set();
+  let result = [];
+  // The initial solution was a nested loop, with a time complexity of O(n**2). This was optimised by using a set to track seen items, thereby reducing the complexity to O(n)
+  for (const item of inputSequence) {
+    if (!seenItems.has(item)) {
+      seenItems.add(item);
+      result.push(item);
     }
   }
-
-  return uniqueItems;
+  return result;
 }
diff --git a/Sprint-2/implement_linked_list/linked_list.py b/Sprint-2/implement_linked_list/linked_list.py
index e69de29b..8375940d 100644
--- a/Sprint-2/implement_linked_list/linked_list.py
+++ b/Sprint-2/implement_linked_list/linked_list.py
@@ -0,0 +1,108 @@
+class Node:
+    """
+    A node in a double linked list.
+
+    Each node supports a data, previous, and next
+    - data: the value
+    - previous: pointer to the previous node
+    - next: pointer to the next node
+    """
+
+    def __init__(self, data):
+        self.data = data
+        self.previous = None
+        self.next = None
+
+class LinkedList:
+    """
+    Double Linked List
+
+    Supports O(1) operations on insertion at head, removal from tail, and removal from arbitrary node
+    """
+    def __init__(self):
+        # Head points to the first node in the list
+        self.head = None
+        
+        # Tail points to the last node in the list
+        self.tail = None
+
+    def is_empty(self):
+        """Returns True if the list contains no nodes."""
+        return self.head is None
+
+    def push_head(self, data):
+        """
+        Insert a new node at the head of the list.
+
+        Time Complexity: O(1)
+
+        Returns:
+            Node: reference to the inserted node (handle for later removal)
+        """
+        node = Node(data)
+
+        if self.is_empty():
+            # If list is empty, head and tail both point to new node
+            self.head = node
+            self.tail = node
+        else:
+            # Link new node before current head
+            node.next = self.head
+            self.head.previous = node
+
+            # Update head pointer
+            self.head = node
+
+        return node
+
+    def pop_tail(self):
+        """
+        Remove and return the value at the tail of the list.
+
+        Time Complexity: O(1)
+
+        Raises:
+            ValueError: if the list is empty
+        """
+
+        if self.is_empty():
+            raise ValueError("List is empty.")
+        
+        # store value before removing node
+        value = self.tail.data
+
+        if self.head == self.tail:
+            # Only one element in the list
+            self.head = None
+            self.tail = None
+
+        else:
+            # Move tail pointer backwards
+            self.tail = self.tail.previous
+
+            # Disconnect old tail node
+            self.tail.next = None
+
+        return value
+
+    def remove(self, node):
+        """
+        Remove a node from the list using its reference.
+
+        Time Complexity: O(1)
+
+        Args:
+            node (Node): the node to remove
+        """
+        if node.previous:
+            node.previous.next = node.next
+        else:
+            self.head = node.next
+
+        if node.next:
+            node.next.previous = node.previous
+        else:
+            self.tail = node.previous
+
+        node.previous = None
+        node.next = None