Time - Yesenia

lib/exercises.rb

@@ -1,19 +1,67 @@
 
 # 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)
+# Space Complexity: O(n)
 
 def grouped_anagrams(strings)
-  raise NotImplementedError, "Method hasn't been implemented yet!"
+  hash = {}
+
+  strings.each do |word|
+    letter_grouping = word.split('').sort().join()
+
+    if !hash[letter_grouping]
+      hash[letter_grouping] = []
+    end 
+
+    hash[letter_grouping] << word
+  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)
+# Space Complexity: O(n)
 def top_k_frequent_elements(list, k)
-  raise NotImplementedError, "Method hasn't been implemented yet!"
+  return [] if list.empty?
+
+  counts_hash = {}
+
+  list.each do |number|
+    if !counts_hash[number]
+      counts_hash[number] = 1 
+    else
+      counts_hash[number] = counts_hash[number] + 1
+    end
+  end
+
+  length_to_num_hash = {}
+
+  counts_hash.keys.each do |number|
+    if !length_to_num_hash[counts_hash[number]]
+      length_to_num_hash[counts_hash[number]] = []
+    end
+    length_to_num_hash[counts_hash[number]] << number
+  end
+
+  result_length = k
+  result = []
+  most_freq_num = length_to_num_hash.keys.max
+
+  while result_length > 0 && most_freq_num != nil && most_freq_num >= 0
+    if length_to_num_hash[most_freq_num]
+      length_to_num_hash[most_freq_num].each do |num|
+        result << num
+        result_length -= 1
+        break if result_length <= 0
+      end
+    end
+    most_freq_num -= 1
+  end
+
+  return result
 end