Earth - Christina Minh

lib/exercises.rb

@@ -1,19 +1,52 @@
 
 # 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 * m * log m) 
+	#	where n is the number of strings in the array to iterate over, 
+	# and m is the number of characters to sort in a string (m log m), 
+# Space Complexity: O(n * m) from sorting characters (m) and returning a new array (n)
+# where n is the number of strings in the array
+# and m is the number of characters in a string 
 def grouped_anagrams(strings)
-  raise NotImplementedError, "Method hasn't been implemented yet!"
+  sorted_strings_hash = Hash.new([])  
+
+  strings.each do |word|
+    word.chars.sort.join
+
+    if sorted_strings_hash[word]
+      sorted_strings_hash[word].push(word)
+		end
+	end
+
+	puts sorted_strings_hash.values
+	return sorted_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(n + m log m) where n is the number of elements in the list and 
+	# and m is the number of unique elements to sort (m log m)
+# Space Complexity: O(m * k) where m is the number of unique elements to put in hash
+	# and k is the number of elements to return
 def top_k_frequent_elements(list, k)
-  raise NotImplementedError, "Method hasn't been implemented yet!"
+	return [] if list.empty?
+
+	frequency_hash = Hash.new(0)
+
+	list.each do |num|
+		frequency_hash[num] += 1
+	end
+
+	sorted_frequency_hash = frequency_hash.sort_by { |num, frequency| -frequency }
+
+	k_frequent_elements = []
+
+	k.times do |i|
+		num_frequency_pair = sorted_frequency_hash[i]
+		k_frequent_elements << num_frequency_pair[0]
+	end
+
+	return k_frequent_elements
 end
 
 
@@ -22,8 +55,54 @@ def top_k_frequent_elements(list, k)
 #   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) constant time to iterate over table that is always 9x9
+# Space Complexity: O(1)
 def valid_sudoku(table)
-  raise NotImplementedError, "Method hasn't been implemented yet!"
-end
+	seen_in_rows = []
+	seen_in_columns = []
+
+	# array index represents the row/column number
+	10.times do |i|
+		seen_in_rows[i] = {}
+		seen_in_columns[i] = {}
+	end
+	# Example seen_in_rows  = [
+	# 	{"5"=>true, "3"=>true, "7"=>true},    					row 0
+	# 	{"6"=>true, "1"=>true, "9"=>true, "5"=>true}, 	row 1
+	# 	{"9"=>true, "8"=>true, "6"=>true}, 							row 2
+	# 	...
+	# ]
+
+	seen_in_subgrids = 
+	[
+		[{}, {}, {}],
+		[{}, {}, {}],
+		[{}, {}, {}],
+	]
+
+	num_rows = table.length
+	num_columns = table[0].length
+
+	num_rows.times do |row|
+		num_columns.times do |col|
+			current_cell_value = table[row][col]
+
+			next if current_cell_value == "."
+
+			# find position of hash in seen_in_subgrids
+			subgrid_col = col / 3
+			subgrid_row = row / 3
+
+			if seen_in_rows[row][current_cell_value] || seen_in_columns[col][current_cell_value] || seen_in_subgrids[subgrid_row][subgrid_col][current_cell_value]
+				return false
+			end	
+
+			# Add value to all look-up hashes
+			seen_in_rows[row][current_cell_value] = true
+			seen_in_columns[col][current_cell_value] = true
+			seen_in_subgrids[subgrid_row][subgrid_col][current_cell_value] = 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 = [