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Earth - Beauttie #14

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120 changes: 111 additions & 9 deletions lib/exercises.rb
Original file line number Diff line number Diff line change
@@ -1,19 +1,64 @@

# 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)

def grouped_anagrams(strings)

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👍 Interesting solution with char.ord This works if you know that all the letters are a-z and roughly the same length. That's because you could end up with two strings of different lengths that have the same sum of characters.

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 +67,65 @@ 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