Ruby way to Check for string palindrome - ruby-on-rails

Question

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

Answer 1

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

Answer 2

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

Answer 3

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.

Answer 4

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)

Answer 5

class String
def palindrome?
self.downcase == self.reverse.downcase
end
end
puts "racecar".palindrome? # true
puts "Racecar".palindrome? # true
puts "mississippi".palindrome? # false

Answer 6

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

Answer 7

> 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

Answer 8

def palindrome?(string)
string[0] == string[-1] && (string.length <= 2 || palindrome?(string[1..-2]))
end

Answer 9

**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

Answer 10

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