Space - Cathy

lib/exercises.rb

@@ -1,19 +1,55 @@
 
 # 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^2) where n is the number of strings? || O(n) where n is the total number of characters? 
+# Space Complexity: O(n) where a new hash is storing the n # of strings
 
 def grouped_anagrams(strings)
-  raise NotImplementedError, "Method hasn't been implemented yet!"
+  return [] if strings.empty?
+  groups = Hash.new()
+
+  # sort each string so we can compare if they look the same == anagrams
+  strings.each do |string|
+    sorted_str = string.split("").sort.join
+
+    # store data into groups
+    if groups[sorted_str]
+      groups[sorted_str] << string
+    else
+      groups[sorted_str] = []
+      groups[sorted_str] << string
+    end
+  end
+  # print groups
+
+  return groups.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) since I'm calling a sort_by enumerable
+# Space Complexity: O(n)
 def top_k_frequent_elements(list, k)
-  raise NotImplementedError, "Method hasn't been implemented yet!"
+  # loop through list and create hash: num => count
+  # then search for the k-most frequent values mapped to key
+  return [] if list.empty?
+  count = {}
+  list.each do |num|
+    count[num] ? count[num] += 1 : count[num] = 1
+  end
+
+  sorted_hash = count.sort_by {|k, v| -v}   # hash#sort_by returns a matrix (2-D array)
+  ans = []
+  i = 0
+  while i < k
+    ans << sorted_hash[i][0]
+    i += 1
+  end
+
+  return ans
+  # ans = count.keys.max_by(k){|key| count[key] }
+  # https://apidock.com/ruby/Enumerable/max_by  # WHY?! what does the max_by actually do?!
 end
 
 
@@ -22,8 +58,81 @@ 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: ?
+# Reference: https://www.tutorialspoint.com/valid-sudoku-in-python
+# Time Complexity: O(n^2) where n is the length of the input table and since table is a 2d array
+# Space Complexity: O(1) where the extra hashes are constant tracking 1~9
 def valid_sudoku(table)
-  raise NotImplementedError, "Method hasn't been implemented yet!"
+  # 1. check rows 2. check columns 3. check sub-boxes
+  # utilize the indices in the 2D-array and use hashed to keep track
+  (0...9).each do |i|
+    row = {}
+    col = {}
+    sub_box = {}
+    # below marks the anchor points for each sub-box: [0,0], [0,3], [0,6], [3,0], [3,3], [3,6], [6,0], [6,3], [6,6]
+    sub_row = 3 * (i / 3)
+    sub_col = 3 * (i % 3)
+
+    (0...9).each do |j|
+      # check rows
+      if table[i][j] =~ /\d/ && row[table[i][j]]
+        return false
+      else
+        row[table[i][j]] = 1
+      end
+      # check columns
+      if table[j][i] =~ /\d/ && col[table[j][i]]
+        return false
+      else
+        col[table[j][i]] = 1
+      end
+      # check sub-boxes: scope the boundaries of sub-boxes according to the anchor points sub_row and sub_col
+      sr = sub_row + (j / 3)
+      sc = sub_col + (j % 3)
+      if table[sr][sc] =~ /\d/ && sub_box[table[sr][sc]]
+        return false
+      else
+        sub_box[table[sr][sc]] = 1
+      end
+    end
+  end
+
+  return true
 end
+
+
+# Thought I could use a hash function to solve grouped_anagrams, but realized the sum/math would cause incorrect results
+# def grouped_anagrams(strings)
+#   # hash function and add sum
+#   # 2 factors: score by alphabet and string length  # can I assume they are anagrams if both factors meet?
+#   return [] if strings.empty?
+#   alpha = {
+#     "a" => 1, "b" => 2, "c" => 3, "d" => 4, "e" => 5, 
+#     "f" => 6, "g" => 7, "h" => 8, "i" => 9, "j" => 10,
+#     "k" => 11, "l" => 12, "m" => 13, "n" => 14, "o" => 15, 
+#     "p" => 16, "q" => 17, "r" => 18, "s" => 19, "t" => 20,
+#     "u" => 21, "v" => 22, "w" => 23, "x" => 24, "y" => 25, "z" => 26,
+#   }
+#   groups = Hash.new()
+
+#   strings.each do |string|
+#     # loop through each letter and count total score # get length
+#     l = 0
+#     sum = 0
+#     while l < string.length
+#       char = string[l]
+#       sum += alpha[char]
+#       l += 1
+#     end
+
+#     # store data into groups
+#     if groups[sum]
+#       groups[sum] << string
+#     else
+#       groups[sum] = []
+#       groups[sum] << string
+#     end
+#   end
+#   # print groups
+
+#   return groups.values
+# end

test/exercises_test.rb

@@ -131,7 +131,7 @@
 
   end
 
-  xdescribe "valid sudoku" do
+  describe "valid sudoku" do
     it "works for the table given in the README" do
       # Arrange
       table = [