diff --git a/hash_practice/exercises.py b/hash_practice/exercises.py
index 48bf95e..5826773 100644
--- a/hash_practice/exercises.py
+++ b/hash_practice/exercises.py
@@ -1,19 +1,43 @@
 
+from ntpath import join
+from sqlalchemy import true
+
+
 def grouped_anagrams(strings):
     """ This method will return an array of arrays.
         Each subarray will have strings which are anagrams of each other
-        Time Complexity: ?
-        Space Complexity: ?
+        Time Complexity: O(n * m log m)
+        ^ Not sure on the time complexity. Here, n represents the number of words in the 
+        input string, while m represents the length of an individual word. 
+        Space Complexity: O(n)
     """
-    pass
+    word_dict = {}
+    for word in strings:
+        sorted_word = ''.join(sorted(word))
+        if sorted_word in word_dict:
+            word_dict[sorted_word].append(word)
+        else:
+            word_dict[sorted_word] = [word]
+    result = [v for k,v in word_dict.items()]
+    return result
+
 
 def top_k_frequent_elements(nums, k):
     """ This method will return the k most common elements
         In the case of a tie it will select the first occuring element.
-        Time Complexity: ?
-        Space Complexity: ?
+        Time Complexity: O(n log n) (This is the result of using the list sort method and
+        n is the number of elements in nums)
+        Space Complexity: O(n)
     """
-    pass
+    if not nums or k < 1:
+        return []
+    freq_map = {}
+    for num in nums:
+        freq_map[num] = 1 + freq_map.get(num, 0)
+    freq_list = [(num, freq) for num, freq in freq_map.items()]
+    freq_list.sort(key = lambda x: x[1], reverse=True)
+    top_el = [freq_list[i][0] for i in range(k)]
+    return top_el
 
 
 def valid_sudoku(table):