According to this tutorial, I only need one end statement for the if, elsif, and else loop:
def dasherize_number(num)
nums = num.to_s
final_string = ''
i = 0
while i < nums.length
if i == 0
final_string += nums[i]
if nums[i].to_i % 2 == 1
final_string += '-'
end
elsif i == nums.length
final_string += nums[i]
else
if i % 2 == 1
final_string += '-'
final_string += nums[i]
end
end
i += 1
end
puts final_string
end
Why do there need to be two end statement between i += 1?
Indentation is important:
def dasherize_number(num)
nums = num.to_s
final_string = ''
i = 0
while i < nums.length
if i == 0
final_string += nums[i]
if nums[i].to_i % 2 == 1
final_string += '-'
end
elsif i == nums.length
final_string += nums[i]
else
if i % 2 == 1
final_string += '-'
final_string += nums[i]
end
end
i += 1
end
puts final_string
end
Related
I have written a code for the drawing board for the minesweeper game, can anyone help me to refactor this code more.
Please find my code below
def draw(height, width, mines)
board = Array.new(height) { Array.new(width,0) }
x = Random.rand(height)
y = Random.rand(width)
mines.times do
until board[x][y] != 'x'
x = Random.rand(height)
y = Random.rand(width)
end
board[x][y] = 'x'
end
board.each_with_index do |row, i|
row.each_with_index do |elem, j|
next if board[i][j] == 'x'
count = 0
count += 1 if i+1 < height && board[i+1][j] == 'x'
count += 1 if j+1 < width && board[i][j+1] == 'x'
count += 1 if i-1 >= 0 && board[i-1][j] == 'x'
count += 1 if j-1 >= 0 && board[i][j-1] == 'x'
board[i][j] = count
end
end
board.each do |row|
row.each do |e|
print "#{e} "
end
print "\n"
end
end
draw(4,4,3)
Thanks in advance.
I think you need to check 8 adjacent cells, not 4. Since this is a refactoring, I kept the original behavior.
def draw(height, width, mines)
board = Array.new(height) { Array.new(width, 0) }
mines.times do
x = rand(height)
y = rand(width)
redo if board[x][y] == 'x'
board[x][y] = 'x'
[[x - 1, y], [x + 1, y], [x, y - 1], [x, y + 1]].each do |x, y|
next if x < 0 || x >= height
next if y < 0 || y >= width
next if board[x][y] == 'x'
board[x][y] += 1
end
end
board.each { |row| puts row.join(' ') }
end
Invoice numbers are numeric only with any number of digits. To format one correctly, group the digits in group of three plus a group of any remainder, but never leave one digit by itself, unless it's a one digit number. Eg these are all correct formatting
123
12-34
6
783-907-23-45
And these are not
123-4
98-456
There's one more catch user input is passed directly to the function and you never know what characters users might type. Ignore any part of the input that is not digit
Invoice.format_number should always return a string
module Invoice
def self.format_number(str)
return ""
end
end
puts Invoice.format_number("ab1234")
What I have tried
1st approach
arr = []
str.chars.each do |elem|
val = elem =~ /\A[-+]?[0-9]*\.?[0-9]+\Z/
arr << elem if val == 0
end
num_of_digits = arr.length
pairs_of_two = 0
pairs_of_three = 0
if num_of_digits > 5
while num_of_digits > 0 do
break if num_of_digits <= 3
if num_of_digits >= 3 && (num_of_digits % 3 == 0 || num_of_digits % 3 == 2)
pairs_of_three += 1
num_of_digits -= 3
elsif num_of_digits % 2 == 0 || num_of_digits % 2 == 1
pairs_of_two += 1
num_of_digits -= 2
end
end
end
2nd approach
arr = []
str.chars.each do |elem|
val = elem =~ /\A[-+]?[0-9]*\.?[0-9]+\Z/
arr << elem if val == 0
end
len = arr.length - 1
if arr.length > 4
str = ""
i = 0
while i < len do
if arr[i..i+3].length == 4
str << arr[i..i+2].join + "-"
i += 3
elsif arr[i..i+2].length == 3
str << arr[i..i+1].join + "-"
i += 2
elsif arr[i..i+1].length == 2
str << arr[i..i+1].join
i += 2
elsif !arr[i].nil?
str << arr[i]
i += 1
end
end
puts str
else
if arr.length <= 3
puts arr.join
else
puts arr[0..1].join + "-" + arr[2..3].join
end
end
But none of them is correct
Here is the function invoice_number in python
def invoice_number(invoice):
s = ''.join(x for x in invoice if x <= '9' and x >= '0')
n = len(s)
if n <= 3:
return s
w = ''
i = 0
while i + 3 <= n:
for j in range(0, 3):
w += s[i + j]
i += 3
w += ('-')
m = n - i
if m == 0: return w[:-1]
if m == 1: return w[:m-3] + '-' + s[-2:]
return w + s[i:]
Testing
print(invoice_number('1234567'))
print(invoice_number('12345678'))
print(invoice_number('abc123456789'))
print(invoice_number('1234abc5678xyz9foobar'))
123-45-67
123-456-78
123-456-789
123-456-789
Eliminating non-digits is easy with re. For your format, the key is to figure our the "right" splitting indices.
Here is a try:
import re
def splits(n, k):
idx = [(i, min(n, i+k)) for i in range(0, n, k)]
if len(idx) > 1:
(a, b), (c, d) = idx[-2:]
if d - c < 2:
idx[-2:] = [(a, b - 1), (c - 1, d)]
return idx
def myformat(s):
s = re.sub(r'[^0-9]+', '', s)
parts = [s[a:b] for a, b in splits(len(s), 3)]
return '-'.join(parts)
Tests:
>>> myformat('123')
123
>>> myformat('1234')
12-34
>>> myformat('6')
6
>>> myformat('7839072345')
783-907-23-45
As the question was asked for ruby, adding solution for ruby. (The inspiration of the code is mostly from #yuri answer)
def format_invoice(invoice)
# only numbers are allowed
invoice = invoice.tr("^0-9","")
#puts invoice
return invoice if(invoice.length <= 3)
formatted_invoice = ''
i = 0
# Loop to divide the invoice in group of 3
while i + 3 <= invoice.length do
for j in 0..2 do
formatted_invoice += invoice[i + j]
end
i += 3
formatted_invoice += ('-')
end
m = invoice.length - i
return formatted_invoice[0..-2] if m == 0
return formatted_invoice[0..m-4] + '-' + invoice[-2..-1] if m == 1
return formatted_invoice + invoice[i..-1]
end
Testing
puts format_invoice('abc1') # 1
puts format_invoice('abc123') # 123
puts format_invoice('abc123A4') # 12-34
puts format_invoice('1234567') # 123-45-67
puts format_invoice('12345678') # 123-456-78
puts format_invoice('abc123456789') # 123-456-789
puts format_invoice('1234a#c5678xyz9foobar') # 123-456-789
I have written the logic for the program to perform FizzBuzz operations:
fizzbuzz
module FizzBuzz
class Operation
def input
puts 'Enter a number upto which Fizz/Buzz needs to be printed'
num = gets.chomp.to_i
fizzbuzz_function(num)
end
def fizzbuzz_function(num)
for i in 1..num
if i % 3 == 0 && i % 5 == 0
puts 'FizzBuzz'
elsif i % 3 == 0
puts 'Fizz'
elsif i % 5 == 0
puts 'Buzz'
else
puts i
end
end
end
end
res = Operation.new
res.input
end
But I am trying to print the output in form of a table.
Here is FizzBuzz in form of a table:
def fizzbuzz_gen(num)
Enumerator.new do |y|
(1..num).each do |i|
if i % 3 == 0 && i % 5 == 0
y << 'FizzBuzz'
elsif i % 3 == 0
y << 'Fizz'
elsif i % 5 == 0
y << 'Buzz'
else
y << i.to_s
end
end
end
end
def fill_to_width(width, e)
result = ""
future_length = -1
while result.length + future_length < width
result << e.next
result << " "
future_length = e.peek.length
end
result.center(width)
end
def format_table(num)
fb = fizzbuzz_gen(num)
begin
puts fill_to_width(75, fb)
puts fill_to_width(75, fb)
loop do
puts "%10s%s%31s%s" % ["", fill_to_width(12, fb), "", fill_to_width(12, fb)]
end
rescue StopIteration
end
end
format_table(100)
There may be less numbers output than specified, in order for one leg not to be shorter than another.
I've just started learning ruby, and the position of where variables are defined somewhat elude me. For example, why does this code work:
def two_sum(nums)
result = nil
i = 0
while i < nums.length
k = (nums.length - 1)
if nums[i] + nums[k] == 0
result = [i,k]
end
i += 1
k -= 1
end
return result
end
And why does this code not work:
def two_sum(nums)
result = nil
i = 0
k = (nums.length - 1)
while i < nums.length
if nums[i] + nums[k] == 0
result = [i,k]
end
i += 1
k -= 1
end
return result
end
Thank you in advance!
I think you code might just have a bug
while i < nums.length
k = (nums.length - 1)
...
k -= 1 # this statement has no effect!
end
Above, the value if k is always (nums.length - 1) because you reassign it at the begin of each iteration. The other statement has no effect.
k = (nums.length - 1)
while i < nums.length
...
k -= 1
end
Above, the value of k starts at (nums.length - 1) in the first iteration and is then reduced by 1 for each iteration.
Pro tipp —
It is very unusual in Ruby to use a for/while/until loop. If you want to loop over all elements use each or each_with_index instead
array.each { |each| ... }
array.each_with_index { |each, n| ... }
I'm not sure why my code is not working, I think my logic is right?
Have the function ExOh(str) take the str parameter being passed and return the string true if there is an equal number of x's and o's, otherwise return the string false. Only these two letters will be entered in the string, no punctuation or numbers. For example: if str is "xooxxxxooxo" then the output should return false because there are 6 x's and 5 o's.
ExOh(str)
i = 0
length = str.length
count_x = 0
count_o = 0
while i < length
if str[i] == "x"
count_x += 1
elsif str[i] == "o"
count_o += 1
end
i+=1
end
if (count_o == count_x)
true
elsif (count_o != count_x)
false
end
end
The problem with your code is the function declaration. Use def ExOh(str) at the start. It may help if you indented also.
def ExOh(str)
i = 0
length = str.length
count_x = 0
count_o = 0
while i < length
if str[i] == "x"
count_x += 1
elsif str[i] == "o"
count_o += 1
end
i+=1
end
if (count_o == count_x)
true
elsif (count_o != count_x)
false
end
end
By the way, a simpler solution using the standard library #count https://ruby-doc.org/core-2.2.0/String.html#method-i-count
def ExOh(str)
str.count('x') == str.count('o')
end