🌊 - Kim

README.md

@@ -39,11 +39,10 @@ Given a non-empty array of integers, return the *k* most frequent elements.
 ```
 Input: nums = [1,1,1,2,2,3], k = 2
 Output: [1,2]
-
 ```
 
 ## Example 2
-
+```
 Input: nums = [1], k = 1
 Output: [1]
 ```

lib/exercises.rb

@@ -1,29 +1,85 @@
 
 # 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) where n is the number of chars
+# Space Complexity: O(n) where n is the number of words
 def grouped_anagrams(strings)
-  raise NotImplementedError, "Method hasn't been implemented yet!"
+  frequency_hash_to_words = Hash.new
+
+  strings.each do |word|
+    frequency_hash = word.each_char.with_object(Hash.new(0)) { |letter, count| count[letter] += 1 }
+
+    if frequency_hash_to_words.include?(frequency_hash)
+      frequency_hash_to_words[frequency_hash] << word
+    else
+      frequency_hash_to_words[frequency_hash] = [ word ]
+    end
+  end
+
+  return frequency_hash_to_words.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 log n)
+# Space Complexity: O(n)
 def top_k_frequent_elements(list, k)
-  raise NotImplementedError, "Method hasn't been implemented yet!"
+  count = Hash.new(0) 
+
+  list.each do |num|
+    count[num] += 1
+  end
+
+  count = count.sort_by { |k, v| -v }.to_h
+
+  return count.keys[0...k]
 end
 
+def get_sub_box(row, col)
+  return 0 if row <= 2 && col <= 2
+  return 1 if row <= 2 && col >= 3 && col <= 5
+  return 2 if row <= 2 && col >= 6
+
+  return 3 if row >= 3 && row <= 5 && col <= 2
+  return 4 if row >= 3 && row <= 5 && col >= 3 && col <= 5
+  return 5 if row >= 3 && row <= 5 && col >= 6
+
+  return 6 if row >= 6 && col <= 2
+  return 7 if row >= 6 && col >= 3 && col <= 5
+  return 8
+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 
 #   row, column or 3x3 subgrid
-# Time Complexity: ?
-# Space Complexity: ?
+# Time Complexity: O(1)
+# Space Complexity: O(n)
 def valid_sudoku(table)
-  raise NotImplementedError, "Method hasn't been implemented yet!"
+  row_number_frequencies = [ {}, {}, {}, {}, {}, {}, {}, {}, {} ]
+  column_number_frequencies = [ {}, {}, {}, {}, {}, {}, {}, {}, {} ]
+  sub_box_number_frequencies = [ {}, {}, {}, {}, {}, {}, {}, {}, {} ]
+
+  table.each_with_index do |row, r|
+    row.each_with_index do |column, c|
+      cell = table[r][c]
+      next if cell == '.'
+
+      # check/update row
+      return false if row_number_frequencies[r].include?(cell)
+      row_number_frequencies[r][cell] = true
+
+      # check/update column
+      return false if column_number_frequencies[c].include?(cell)
+      column_number_frequencies[c][cell] = true
+
+      # check/update sub-box
+      sub_box = get_sub_box(r, c)
+      return false if sub_box_number_frequencies[sub_box].include?(cell)
+      sub_box_number_frequencies[sub_box][cell] = true
+    end
+  end
+
+  return true
 end

test/exercises_test.rb

@@ -151,7 +151,7 @@
     end
   end
 
-  xdescribe "valid sudoku" do
+  describe "valid sudoku" do
     it "works for the table given in the README" do
       # Arrange
       table = [