-
Notifications
You must be signed in to change notification settings - Fork 39
New issue
Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.
By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.
Already on GitHub? Sign in to your account
Earth - Beauttie #14
base: master
Are you sure you want to change the base?
Earth - Beauttie #14
Changes from 7 commits
4b5ab7c
a859e23
58c5649
c14d6ab
d39785b
e0152f7
e3aa466
c28510a
File filter
Filter by extension
Conversations
Jump to
Diff view
Diff view
There are no files selected for viewing
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) | ||
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
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 👍 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 | ||
|
||
|
||
|
@@ -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) | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 👍 |
||
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) | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 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 |
There was a problem hiding this comment.
Choose a reason for hiding this comment
The reason will be displayed to describe this comment to others. Learn more.
👍 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.