diff --git a/lib/exercises.rb b/lib/exercises.rb
index e1b3850..5085018 100644
--- a/lib/exercises.rb
+++ b/lib/exercises.rb
@@ -1,19 +1,44 @@
 
 # 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) - Nested loop
+# Space Complexity: O(n)
 
 def grouped_anagrams(strings)
-  raise NotImplementedError, "Method hasn't been implemented yet!"
+  anagram_hash = {}
+
+  strings.each do |str|
+    bucket = anagram_hash.keys.find do |b|
+      unique_chars = b.chars.sort.join
+      unique_chars == str.chars.sort.join
+    end
+
+    if !bucket.nil?
+      anagram_hash[bucket] << str
+    else
+      anagram_hash[str] = [str]
+    end
+  end
+  return anagram_hash.values
 end
 
 # 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)
+# Space Complexity: O(n)
 def top_k_frequent_elements(list, k)
-  raise NotImplementedError, "Method hasn't been implemented yet!"
+  elements_hash = {}
+
+  list.each do |element|
+    if elements_hash[element]
+      elements_hash[element] += 1
+    else
+      elements_hash[element] = 1
+    end
+  end
+
+  #this doesn't pass tests — instead of grabbing the FIRST most frequent, it grabs the most frequent, period. I couldn't figure out how to do the other way
+  return elements_hash.keys.max_by(k) { |key| elements_hash[key] }
 end
 
 
@@ -25,5 +50,5 @@ def top_k_frequent_elements(list, k)
 # Time Complexity: ?
 # Space Complexity: ?
 def valid_sudoku(table)
-  raise NotImplementedError, "Method hasn't been implemented yet!"
+  # raise NotImplementedError, "Method hasn't been implemented yet!"
 end