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Earth - Beauttie #14

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123 changes: 114 additions & 9 deletions lib/exercises.rb
Original file line number Diff line number Diff line change
@@ -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^2)
# Space Complexity: O(n)

# sort the letters in the string alphabetically
# and use that sorted string as the key => O(n log n)
# or use a hash containing letter counts as the key
def grouped_anagrams(strings)
raise NotImplementedError, "Method hasn't been implemented yet!"
hash = Hash.new
strings.each do |string|
if hash[str_to_int(string)]
hash[str_to_int(string)] << string
else
hash[str_to_int(string)] = [string]
end
end

return hash.values
end

def str_to_int(string)
return string.chars.map { |char| char.ord }.sum
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)
Comment on lines +29 to 31

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👍 I like the O(n) solution here.

raise NotImplementedError, "Method hasn't been implemented yet!"
return list if list.empty? || list.length == 1

hash = Hash.new(0)
list.each { |int| hash[int] += 1 }

max_count = 0
inverted_hash = Hash.new
hash.each do |key, value|
if inverted_hash[value]
inverted_hash[value] << key
else
inverted_hash[value] = [key]
end

max_count = value if value > max_count
end

k_dup = k.dup
k_most_frequent = []
max_count.downto(1) do |count|
if inverted_hash[count]
inverted_hash[count].each do |int|
if k_dup > 0
k_most_frequent << int
k_dup -= 1
else
return k_most_frequent
end
end
end
end

return k_most_frequent
end


Expand All @@ -22,8 +70,65 @@ 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) because the table has nine rows and columns
# Space Complexity: O(1) because each hash will store up to nine cells
def valid_sudoku(table)

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👍

raise NotImplementedError, "Method hasn't been implemented yet!"
return valid_rows(table) && valid_columns(table) && valid_boxes(table)
end

def valid_cell(char)
return char != "." && /[1-9]/.match(char)
end

def valid_rows(table)
hash_row = Hash.new
table.each do |row|
# Each row is an array
row.each do |cell|
return false if hash_row[cell] && valid_cell(cell)
hash_row[cell] = true
end
hash_row.clear
end

return true
end

def valid_columns(table)
hash_column = Hash.new
i = 0
j = 0
while j < table[i].length
# Checks column
while i < table.length
return false if hash_column[table[i][j]] && valid_cell(table[i][j])
hash_column[table[i][j]] = true
i += 1
end
hash_column.clear
# Reset row number back to zero
i = 0
j += 1
end

return true
end

def valid_boxes(table)

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I like the helper methods here.

hash_box = Hash.new
3.times do |row_corner|
3.times do |col_corner|
3.times do |row|
3.times do |col|
box_row = row + 3 * row_corner
box_col = col + 3 * col_corner
return false if hash_box[table[box_row][box_col]] && valid_cell(table[box_row][box_col])
hash_box[table[box_row][box_col]] = true
end
end
hash_box.clear
end
end

return true
end
2 changes: 1 addition & 1 deletion test/exercises_test.rb
Original file line number Diff line number Diff line change
Expand Up @@ -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 = [
Expand Down