diff --git a/lib/exercises.rb b/lib/exercises.rb
index e1b3850..4773154 100644
--- a/lib/exercises.rb
+++ b/lib/exercises.rb
@@ -1,19 +1,39 @@
 
 # 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 * mlogm) where n is the number of strings and m is the number of letters in a string
+# Space Complexity: O(n) where n is the number of strings
 
 def grouped_anagrams(strings)
-  raise NotImplementedError, "Method hasn't been implemented yet!"
+  strings_hash = Hash.new{|hash, key| hash[key] = []}
+  strings.each { |string| strings_hash[string.downcase.chars.sort.join] << string}
+  return strings_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(nlogn) (for the sort)
+# Space Complexity: O(n)
 def top_k_frequent_elements(list, k)
-  raise NotImplementedError, "Method hasn't been implemented yet!"
+  return [] if list.empty?
+
+  element_count = Hash.new(0)
+  element_first_occur = Hash.new
+
+  list.each_with_index do |num, i|
+    element_count[num] += 1
+    element_first_occur[num] = i unless element_first_occur[num]
+  end
+
+  sorted = element_count.keys.sort_by{|num| element_count[num]}.reverse
+  last_num_count = element_count[sorted[k - 1]]
+
+  if(last_num_count == element_count[sorted[k]])
+    sorted[k - 1] = sorted
+      .filter{|num| element_count[num] == last_num_count}
+      .min_by{|num| element_first_occur[num]}
+  end
+  sorted[0...k]
 end