Related
I recently solved this problem, but felt there is a simpler way to do it. I looked into inject, step, and map, but couldn't figure out how to implement them into this code. I want to use fewer lines of code than I am now. I'm new to ruby so if the answer is simple I'd love to add it to my toolbag. Thank you in advance.
goal: accept a sentence string as an arg, and return the sentence with words alternating between uppercase and lowercase
def alternating_case(str)
newstr = []
words = str.split
words.each.with_index do |word, i|
if i.even?
newstr << word.upcase
else
newstr << word.downcase
end
end
newstr.join(" ")
end
You could reduce the number of lines in the each_with_index block by using a ternary conditional (true/false ? value_if_true : value_if_false):
words.each.with_index do |word, i|
newstr << i.even? ? word.upcase : word.downcase
end
As for a different way altogether, you could iterate over the initial string, letter-by-letter, and then change the method when you hit a space:
def alternating_case(str)
#downcase = true
new_str = str.map { |letter| set_case(letter)}
end
def set_case(letter)
#downcase != #downcase if letter == ' '
return #downcase ? letter.downcase : letter.upcase
end
We can achieve this by using ruby's Array#cycle.
Array#cycle returns an Enumerator object which calls block for each element of enum repeatedly n times or forever if none or nil is given.
cycle_enum = [:upcase, :downcase].cycle
#=> #<Enumerator: [:upcase, :downcase]:cycle>
5.times.map { cycle_enum.next }
#=> [:upcase, :downcase, :upcase, :downcase, :upcase]
Now, using the above we can write it as following:
word = "dummyword"
cycle_enum = [:upcase, :downcase].cycle
word.chars.map { |c| c.public_send(cycle_enum.next) }.join("")
#=> "DuMmYwOrD"
Note: If you are new to ruby, you may not be familiar with public_send or Enumberable module. You can use the following references.
Enumberable#cycle
#send & #public_send
I have 2 strings:
a = "qwer"
b = "asd"
Result = "qawsedr"
Same is the length of b is greater than a. show alternate the characters.
What is the best way to do this? Should I use loop?
You can get the chars from your a and b string to work with them as arrays and then "merge" them using zip, then join them.
In the case of strings with different length, the array values must be reversed, so:
def merge_alternately(a, b)
a = a.chars
b = b.chars
if a.length >= b.length
a.zip(b)
else
array = b.zip(a)
array.map{|e| e != array[-1] ? e.reverse : e}
end
end
p merge_alternately('abc', 'def').join
# => "adbecf"
p merge_alternately('ab', 'zsd').join
# => "azbsd"
p merge_alternately('qwer', 'asd').join
# => "qawsedr"
Sebastián's answer gets the job done, but it's needlessly complex. Here's an alternative:
def merge_alternately(a, b)
len = [a.size, b.size].max
Array.new(len) {|n| [ a[n], b[n] ] }.join
end
merge_alternately("ab", "zsd")
# => "azbsd"
The first line gets the size of the longer string. The second line uses the block form of the Array constructor; it yields the indexes from 0 to len-1 to the block, resulting in an array like [["a", "z"], ["b", "s"], [nil, "d"]]. join turns it into a string, conveniently calling to_s on each item, which turns nil into "".
Here's another version that does basically the same thing, but skips the intermediate arrays:
def merge_alternately(a, b)
len = [a.size, b.size].max
len.times.reduce("") {|s, i| s + a[i].to_s + b[i].to_s }
end
len.times yields an Enumerator that yields the indexes from 0 to len-1. reduce starts with an empty string s and in each iteration appends the next characters from a and b (or ""—nil.to_s—if a string runs out of characters).
You can see both on repl.it: https://repl.it/I6c8/1
Just for fun, here's a couple more solutions. This one works a lot like Sebastián's solution, but pads the first array of characters with nils if it's shorter than the second:
def merge_alternately(a, b)
a, b = a.chars, b.chars
a[b.size - 1] = nil if a.size < b.size
a.zip(b).join
end
And it wouldn't be a Ruby answer without a little gsub:
def merge_alternately2(a, b)
if a.size < b.size
b.gsub(/./) { a[$`.size].to_s + $& }
else
a.gsub(/./) { $& + b[$`.size].to_s }
end
end
See these two on repl.it: https://repl.it/I6c8/2
I have the following strings:
",||||||||||||||"
",|||||a|||||,|"
I would like to achieve that all occurrences of ",|" are replaced with ",,"
The output should be the following:
",,,,,,,,,,,,,,,"
",,,,,,a|||||,,"
When I run .gsub(',|', ',,') on the strings I get not the desired output.
",,|||||||||||||"
",,||||a|||||,,"
That's because it does not run gsub several times.
Is there a similar method that runs recursively.
A regular expression matches can not overlap. Since matches are what is used for replacement, you can't do it that way. Here's two workarounds:
str = ",|||||a|||||,|"
while str.gsub!(/,\|/, ',,'); end
str = ",|||||a|||||,|"
str.gsub!(/,(\|+)/) { "," * ($1.length + 1) }
smoke_weed_every_day = lambda do |piper|
commatosed = piper.gsub(',|', ',,')
commatosed == piper ? piper : smoke_weed_every_day.(commatosed)
end
smoke_weed_every_day.(",||||||||||||||") # => ",,,,,,,,,,,,,,,"
smoke_weed_every_day.(",|||||a|||||,|") # => ",,,,,,a|||||,,"
From an old library of mine. This method iterates until the block output is equal to its input :
def loop_until_convergence(x)
x = yield(previous = x) until previous == x
x
end
puts loop_until_convergence(',||||||||||||||') { |s| s.gsub(',|', ',,') }
# ",,,,,,,,,,,,,,,"
puts loop_until_convergence(',|||||a|||||,|') { |s| s.gsub(',|', ',,') }
# ",,,,,,a|||||,,"
As a bonus, you can calculate a square root in very few iterations :
def root(n)
loop_until_convergence(1) { |x| 0.5 * (x + n / x) }
end
p root(2)
# 1.414213562373095
p root(3)
# 1.7320508075688772
As with #Amandan's second solution there is no need to iterate until no further changes are made.
COMMA = ','
PIPE = '|'
def replace_pipes_after_comma(str)
run = false
str.gsub(/./) do |s|
case s
when PIPE
run ? COMMA : PIPE
when COMMA
run = true
COMMA
else
run = false
s
end
end
end
replace_pipes_after_comma ",||||||||||||||"
#=> ",,,,,,,,,,,,,,,"
replace_pipes_after_comma ",|||||a|||||,|"
#=> ",,,,,,a|||||,,"
I have a string a5bc2cdf3. I want to expand it to aaaaabcbccdfcdfcdf.
In the string is a5, so the resulting string should contain 5 consecutive "a"s, "bc2" results in "bc" appearing 2 times consecutively, and cdf should repeat 3 times.
If input is a5bc2cdf3, and output is aaaaabcbccdfcdfcdf how can I do this in a Ruby method?
def get_character("compressed_string",index)
expanded_string = calculate_expanded_string(compressed_string)
required_char = expanded_string(char_at, index_number(for eg 3))
end
def calculate_expanded_string(compressed_string)
return expanded
end
You may use a regex like
.gsub(/([a-zA-Z]+)(\d+)/){$1*$2.to_i}
See the Ruby online demo
The /([a-zA-Z]+)(\d+)/ will match stubstrings with 1+ letters (([a-zA-Z]+)) and 1+ digits ((\d+)) and will capture them into 2 groups that are later used inside a block to return the string you need.
Note that instead of [a-zA-Z] you might consider using \p{L} that can match any letters.
You want to break out of gsub once the specified index is reached in the original "compressed" string. It is still possible, see this Ruby demo:
s = 'a5bc2cdf3' # input string
index = 5 # break index
result = "" # expanded string
s.gsub!(/([a-zA-Z]+)(\d+)/){ # regex replacement
result << $1*$2.to_i # add to the resulting string
break if Regexp.last_match.end(0) >= index # Break if the current match end index is bigger or equal to index
}
puts result[index] # Show the result
# => b
For brevity, you may replace Regexp.last_match with $~.
I would propose to use scan to move over the compressed string, using a simple RegEx which detects groups of non-decimal characters followed by their count as decimal /([^\d]+)(\d+)/.
def get_character(compressed_string, index)
result = nil
compressed_string.scan(/([^\d]+)(\d+)/).inject(0) do |total_length, (chars, count)|
decoded_string = chars * count.to_i
total_length += decoded_string.length
if index < total_length
result = decoded_string[-(total_length - index)]
break
else
total_length
end
end
result
end
Knowing the current (total) length, one can break out of the loop if the current expanded string includes the requested index. The string is never decoded entirely.
This code gives the following results
get_character("a5bc2cdf3", 5) # => "b"
get_character("a5bc2cdf3", 10) # => "d"
get_character("a5bc2cdf3", 20) # => nil
Just another way. I prefer Wiktor's method by a long way.
def stringy str, index
lets, nums = str.split(/\d+/), str.split(/[a-z]+/)[1..-1].map(&:to_i)
ostr = lets.zip(nums).map { |l,n| l*n }.join
ostr[index]
end
str = 'a5bc2cdf3'
p stringy str, 5 #=> "b"
I'd use:
str = "a5bc2cdf3"
str.split(/(\d+)/).each_slice(2).map { |s, c| s * c.to_i }.join # => "aaaaabcbccdfcdfcdf"
Here's how it breaks down:
str.split(/(\d+)/) # => ["a", "5", "bc", "2", "cdf", "3"]
This works because split will return the value being split on if it's in a regex group: /(\d+)/.
str.split(/(\d+)/).each_slice(2).to_a # => [["a", "5"], ["bc", "2"], ["cdf", "3"]]
The resulting array can be broken into the string to be repeated and its associated count using each_slice(2).
str.split(/(\d+)/).each_slice(2).map { |s, c| s * c.to_i } # => ["aaaaa", "bcbc", "cdfcdfcdf"]
That array of arrays can then be processed in a map that uses String's * to repeat the characters.
And finally join concatenates all the resulting expanded strings back into a single string.
I wanted to check if a string is palindrome or not using ruby code.
I am a starter in ruby so not too aquainted with the string methods in ruby
If you are not acquainted with Ruby's String methods, you should have a look at the documentation, it's very good. Mithun's answer already showed you the basic principle, but since you are new to Ruby, there's a couple more things to keep in mind:
*) If you have a predicate method, it's customary to name it with a trailing question mark, e.g. palindrome?.
*) Boolean expressions evaluate to a boolean, so you don't need to explicitly return true or false. Hence a short idiomatic version would be
def palindrome?(str)
str == str.reverse
end
*) Since Ruby's classes are open, you could add this to the string class:
class String
def palindrome?
self == self.reverse
end
end
*) If you don't want to monkey-patch String, you can directly define the method on single object (or use a module and Object#extend):
foo = "racecar"
def foo.palindrome?
self == self.reverse
end
*) You might want to make the palindrome check a bit more complex, e.g. when it comes to case or whitespace, so you are also able to detect palindromic sentences, capitalized words like "Racecar" etc.
pal = "Never a foot too far, even."
class String
def palindrome?
letters = self.downcase.scan(/\w/)
letters == letters.reverse
end
end
pal.palindrome? #=> true
def check_palindromic(variable)
if variable.reverse == variable #Check if string same when reversed
puts "#{ variable } is a palindrome."
else # If string is not the same when reversed
puts "#{ variable } is not a palindrome."
end
end
The recursive solution shows how strings can be indexed in Ruby:
def palindrome?(string)
if string.length == 1 || string.length == 0
true
else
if string[0] == string[-1]
palindrome?(string[1..-2])
else
false
end
end
end
If reading the Ruby string documentation is too boring for you, try playing around with the Ruby practice questions on CodeQuizzes and you will pick up most of the important methods.
def is_palindrome(value)
value.downcase!
# Reverse the string
reversed = ""
count = value.length
while count > 0
count -= 1
reversed += value[count]
end
# Instead of writing codes for reverse string
# we can also use reverse ruby method
# something like this value == value.reverse
if value == reversed
return "#{value} is a palindrom"
else
return "#{value} is not a palindrom"
end
end
puts "Enter a Word"
a = gets.chomp
p is_palindrome(a)
class String
def palindrome?
self.downcase == self.reverse.downcase
end
end
puts "racecar".palindrome? # true
puts "Racecar".palindrome? # true
puts "mississippi".palindrome? # false
str= gets.chomp
str_rev=""
n=1
while str.length >=n
str_rev+=str[-n]
n+=1
end
if str_rev==str
puts "YES"
else
puts "NO"
end
> first method
a= "malayalam"
if a == a.reverse
puts "a is true"
else
puts "false"
end
> second one
a= "malayalam"
a=a.split("")
i=0
ans=[]
a.count.times do
i=i+1
k=a[-(i)]
ans << k
end
if a== ans
puts "true"
else
puts "false"
end
def palindrome?(string)
string[0] == string[-1] && (string.length <= 2 || palindrome?(string[1..-2]))
end
**Solution 1** Time complexity = O(n), Space complexity = O(n)
This solution does not use the reverse method of the String class. It uses a stack(we could use an array that only allows entry and exit of elements from one end to mimic a stack).
def is_palindrome(str)
stack = []
reversed_str = ''
str.each_char do |char|
stack << char
end
until stack.empty?
reversed_str += stack.pop
end
if reversed_str == str
return true
else
return false
end
end
` Solution 2: Time complexity = O(n), Space complexity = O(1)
def inplace_reversal!(str)
i =0
j = str.length - 1
while i < j
temp = str[i]
str[i] = str[j]
str[j] = temp
i+=1
j-=1
end
return str
end
def palindrome?(str)
return "Please pass the string" if str.nil?
str = str.downcase
str_array = str.split('')
reverse_string = str_array.each_index{ |index| str_array[str_array.count - index - 1 ] end
return ("String #{str} is not a palindrome") unless str == reverse_string.join('')
"String #{str} is palindrome"
end