diff --git a/02_activities/assignments/assignment_1.ipynb b/02_activities/assignments/assignment_1.ipynb
index 625be0241..891829495 100644
--- a/02_activities/assignments/assignment_1.ipynb
+++ b/02_activities/assignments/assignment_1.ipynb
@@ -56,32 +56,101 @@
   },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 12,
    "metadata": {},
-   "outputs": [],
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "True"
+      ]
+     },
+     "execution_count": 12,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
    "source": [
     "# For testing purposes, we will write our code in the function\n",
     "def anagram_checker(word_a, word_b):\n",
     "  # Your code here\n",
+    "  # Dictionary used to count the frequency of each character\n",
+    "  frequency = {}\n",
+    "  # Case-insensitive comparison\n",
+    "  word_a = word_a.lower()\n",
+    "  word_b = word_b.lower()\n",
+    "\n",
+    "  # If the length of word_a is not equal to the length of word_b,\n",
+    "  # the words are definitely not anagrams\n",
+    "  if len(word_a) != len(word_b):\n",
+    "    return False\n",
+    "  \n",
+    "  # Count the occurrence of each character in word_a\n",
+    "  # O(n)\n",
+    "  for char in word_a:\n",
+    "    if char not in frequency:\n",
+    "      frequency[char] = 1\n",
+    "    else:\n",
+    "      frequency[char] += 1\n",
     "\n",
+    "  # If there is a character in word_b that does not exist in word_a,\n",
+    "  # or if a character occurs more times in word_b than in word_a,\n",
+    "  # then the words are not anagrams  \n",
+    "  # O(m)\n",
+    "  for char in word_b:\n",
+    "    if char not in frequency or frequency[char] == 0:\n",
+    "      return False\n",
+    "    else:\n",
+    "      frequency[char] -= 1\n",
+    "  \n",
+    "  # If the frequency of each character in word_a and word_b is exactly the same,\n",
+    "  # then the frequency differences must all be zero\n",
+    "  # O(k)\n",
+    "  for count in frequency.values():\n",
+    "    if count != 0:\n",
+    "      return False\n",
+    "  # O(n) + O(m) + O(k) ~> O(n)\n",
+    "  return True\n",
     "# Run your code to check using the words below:\n",
     "anagram_checker(\"Silent\", \"listen\")"
    ]
   },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 13,
    "metadata": {},
-   "outputs": [],
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "False"
+      ]
+     },
+     "execution_count": 13,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
    "source": [
     "anagram_checker(\"Silent\", \"Night\")"
    ]
   },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 14,
    "metadata": {},
-   "outputs": [],
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "True"
+      ]
+     },
+     "execution_count": 14,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
    "source": [
     "anagram_checker(\"night\", \"Thing\")"
    ]
@@ -103,7 +172,45 @@
    "source": [
     "def anagram_checker(word_a, word_b, is_case_sensitive):\n",
     "  # Modify your existing code here\n",
+    "  # Dictionary used to count the frequency of each character\n",
+    "  frequency = {}\n",
+    "\n",
+    "  if not is_case_sensitive:\n",
+    "    # Case-insensitive comparison\n",
+    "    word_a = word_a.lower()\n",
+    "    word_b = word_b.lower()\n",
+    "\n",
+    "  # If the length of word_a is not equal to the length of word_b,\n",
+    "  # the words are definitely not anagrams\n",
+    "  if len(word_a) != len(word_b):\n",
+    "    return False\n",
+    "  \n",
+    "  # Count the occurrence of each character in word_a\n",
+    "  # O(n)\n",
+    "  for char in word_a:\n",
+    "    if char not in frequency:\n",
+    "      frequency[char] = 1\n",
+    "    else:\n",
+    "      frequency[char] += 1\n",
     "\n",
+    "  # If there is a character in word_b that does not exist in word_a,\n",
+    "  # or if a character occurs more times in word_b than in word_a,\n",
+    "  # then the words are not anagrams  \n",
+    "  # O(m)\n",
+    "  for char in word_b:\n",
+    "    if char not in frequency or frequency[char] == 0:\n",
+    "      return False\n",
+    "    else:\n",
+    "      frequency[char] -= 1\n",
+    "  \n",
+    "  # If the frequency of each character in word_a and word_b is exactly the same,\n",
+    "  # then the frequency differences must all be zero\n",
+    "  # O(k)\n",
+    "  for count in frequency.values():\n",
+    "    if count != 0:\n",
+    "      return False\n",
+    "  # O(n) + O(m) + O(k) ~> O(n)\n",
+    "  return True\n",
     "# Run your code to check using the words below:\n",
     "anagram_checker(\"Silent\", \"listen\", False) # True"
    ]
@@ -130,7 +237,7 @@
  ],
  "metadata": {
   "kernelspec": {
-   "display_name": "new-learner",
+   "display_name": "python-env",
    "language": "python",
    "name": "python3"
   },
@@ -144,7 +251,7 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.11.8"
+   "version": "3.11.7"
   }
  },
  "nbformat": 4,