From 26c842afe6a780a51ef4b090c7b589e50a719b8b Mon Sep 17 00:00:00 2001
From: Alexandria McCarthy <alexandria.mccarthy77@gmail.com>
Date: Sun, 15 Sep 2019 14:02:50 -0700
Subject: [PATCH 1/2] wrote first two methods to make tests pass

---
 lib/exercises.rb       | 61 +++++++++++++++++++++++++++++++++++-------
 test/exercises_test.rb |  2 +-
 2 files changed, 53 insertions(+), 10 deletions(-)

diff --git a/lib/exercises.rb b/lib/exercises.rb
index 2cb2bfa..92ef388 100644
--- a/lib/exercises.rb
+++ b/lib/exercises.rb
@@ -2,29 +2,72 @@
 
 # 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 length of the strings array and m is the length of each word
+#                   since each word is getting sorted
+# Space Complexity: O(n) where n is the size of the strings array - worst case there are no anagrams and the
+#                    hash is the same size as the array
 
 def grouped_anagrams(strings)
-  raise NotImplementedError, "Method hasn't been implemented yet!"
+  return [] if strings.length == 0
+
+  hash = {}
+
+  strings.each do |word|
+    sorted_word = word.chars.sort.join
+
+    if !hash[sorted_word]
+      hash[sorted_word] = [word]
+    else
+      hash[sorted_word] << word
+    end
+  end
+
+  return 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) where n is the length of the list
+# Space Complexity: O(n) where n is the length of the lists
 def top_k_frequent_elements(list, k)
-  raise NotImplementedError, "Method hasn't been implemented yet!"
-end
+  return [] if list.length == 0
+
+  hash = {}
+  list.each do |ele|
+    if !hash[ele]
+      hash[ele] = 1
+    else
+      hash[ele] += 1
+    end
+  end
+
+  top_occurring = []
+
+  k.times do |i|
+    top_val = nil
+    top_freq = 0
 
+    hash.each do |ele, freq|
+      if freq > top_freq
+        top_freq = freq
+        top_val = ele
+      end
+    end
+
+    hash.delete(top_val)
+    top_occurring.push(top_val)
+  end
+
+  return top_occurring
+end
 
 # This method will return the true if the table is still
 #   a valid sudoku table.
 #   Each element can either be a ".", or a digit 1-9
-#   The same digit cannot appear twice or more in the same 
+#   The same digit cannot appear twice or more in the same
 #   row, column or 3x3 subgrid
 # Time Complexity: ?
 # Space Complexity: ?
 def valid_sudoku(table)
   raise NotImplementedError, "Method hasn't been implemented yet!"
-end
\ No newline at end of file
+end
diff --git a/test/exercises_test.rb b/test/exercises_test.rb
index a649110..dd93104 100644
--- a/test/exercises_test.rb
+++ b/test/exercises_test.rb
@@ -68,7 +68,7 @@
     end
   end
 
-  xdescribe "top_k_frequent_elements" do
+  describe "top_k_frequent_elements" do
     it "works with example 1" do
       # Arrange
       list = [1,1,1,2,2,3]

From 8dbf80a3f52891d56078e1dd607430e4ca0b384c Mon Sep 17 00:00:00 2001
From: Alexandria McCarthy <alexandria.mccarthy77@gmail.com>
Date: Thu, 19 Sep 2019 13:35:08 -0700
Subject: [PATCH 2/2] fixed time complexity

---
 lib/exercises.rb | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/lib/exercises.rb b/lib/exercises.rb
index 92ef388..303ffa1 100644
--- a/lib/exercises.rb
+++ b/lib/exercises.rb
@@ -27,7 +27,7 @@ def grouped_anagrams(strings)
 
 # This method will return the k most common elements
 # in the case of a tie it will select the first occuring element.
-# Time Complexity: O(n) where n is the length of the list
+# Time Complexity: O(n * k) where n is the length of the list 
 # Space Complexity: O(n) where n is the length of the lists
 def top_k_frequent_elements(list, k)
   return [] if list.length == 0