diff --git a/hash_practice/exercises.py b/hash_practice/exercises.py
index 48bf95e..4f0edbc 100644
--- a/hash_practice/exercises.py
+++ b/hash_practice/exercises.py
@@ -1,20 +1,98 @@
 
+def create_freq_map(word):
+    freq_map = {}
+    for char in word:
+        if char in freq_map:
+            freq_map[char] += 1
+        else:
+            freq_map[char] = 1
+    # convert to frozenset because it's hashable whereas dicts and sets are not
+    return freq_map
+
 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^2)
+        Space Complexity: O(n)
+
+    Ideas
+    Have freq maps as keys and words as values in a list
+    dict = {
+        {"a":1, "t":1, "e":1}: ["ate", "eat", "tea"]
+    }
+    -> would need 3 hash tables: 1 for hash of hash tables of frequences, 2 dict has keys?
+    -> or create this dict example without creating another hash
+
+    PSEUDO
+    - Create empty dict
+    - Loop through each word in words and create freq map (maybe use a helper)
+    - if freq map not in dict, append to dict with freq map as key and word as value
+    - if freq map in dict, append word to value
+    - return dict values as list
     """
-    pass
+    word_dict = {}
+    for word in strings:
+        current_freq_map = frozenset(create_freq_map(word))
+        if current_freq_map in word_dict:
+            word_dict[current_freq_map] += [word]
+        else:
+            word_dict[current_freq_map] = [word]
+
+    return list(word_dict.values())
 
 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)
+        Space Complexity: O(n)
+
+    Ideas
+    Create freq map where keys are nums and value is the frequency
+    How to return the number with 1st, 2nd highest freq if k =2
+    what if k=3
+
+    PSEUDO
+    -Create freq map 
+    freq_map = {
+        1:1,
+        2:3,
+        3:2
+    }
+    - have a max_count for max value ->
+        - could also maybe use slicing 
+    -create empty list
+    - for i in range(k), append key that has max count
+    - decrement max_count
+
+    with duplicate max values:
+    - create max_list instead of a variable
+    - sort list
+    - slice list
+    - iterate through freq map and add values that match the sliced list elements to a set
+    to avoid duplicates
     """
-    pass
 
+    freq_map = {}
+    for num in nums:
+        if num in freq_map:
+            freq_map[num] += 1
+        else:
+            freq_map[num] = 1
+
+    freq_values = list(freq_map.values())
+    # sorts values of freq map in descending order
+    freq_values.sort(reverse=True)
+    # k number of max frequencies. Assumes if k = 2, and there are more than one with the 2nd highest frequency
+    # ex output = [1,1,2,2,3,3,3]
+    # 1 and 2 are tied so they are both added to max_values and the output will have 3 elements in a list. output = [1,2,3]
+    max_values = freq_values[:k]
+
+    answer = set()
+    for key, value in freq_map.items():
+        for max in max_values:
+            if value == max:
+                answer.add(key)
+    return list(answer)
 
 def valid_sudoku(table):
     """ This method will return the true if the table is still