scramblies.md

Scramblies

Check out the instructions on Codewars

Complete the function scramble(str1, str2) that returns true if a portion of str1 characters can be rearranged to match str2, otherwise returns false.

Solution 1 - Hash

In first solution we could have used with Ruby's inbuilt tally. But we also provided our own tally method because most online IDEs don't run with Ruby 2.7. yet which introduced tally to the Enumerable module.

def scramble(s1, s2)
  mixed = tally(s1)
  word = tally(s2)
  word.all? { |key, value| value <= mixed[key] }
end

# for older Ruby versions, that don't support the native tally
def tally(str)
  str.split(//).reduce(Hash.new(0)) do |tally, char|
    tally[char] += 1
    tally
  end
end

If you have a hard time understanding reduce here is our own tally version with .each instead. When using each as the iterative method we have to create the Hash that saves the occurrences ourselves.

def tally(str)
  tally = Hash.new(0)
  str.split(//).each do |char|
    tally[char] += 1
  end
  tally
end

Solution 2 - String methods

A nice collection of string methods is used in this solution. Each_char makes it easy to iterate over a string without splitting it into an array first.And sub! is used to delete characters from the bigger string.

def scramble(s1, s2)
  s2.each_char do |char|
    return false unless s1.include? char

    s1.sub!(char, '')
  end
  true
end

Solution 3 - Create your own Array#frequency

Here we're adding out own frequency method to the Array Class.

class Array
  def frequency
    Hash.new(0).tap { |counts| each { |v| counts[v] +=1} }
   end
 end

def scramble(s1,s2)

  split1 = s1.chars.frequency
  split2 = s2.chars.frequency

  split2.all? { |char2, count2| split1[char2] >= count2 }
end

Solution 4 - Two pointers

When both strings are sorted you can move over them with two pointers combining each single character as you go. If the characters are the same, or if the longer string has more characters that are earlier in the alphabet. A little example:

string1 = "katas"
string2 = "steak"

sorted1 = "aakst"
sorted2 = "aekst"

👆 Looking at the first two characters - a and a they're both the same, so we can both advance both pointers. So in the second loop we look at the second characters respectively - a and e. Since a comes before e in the alphabet we can assume that the string1 could have an e later on, so no need to cancel the whole mission just yet, we will only advance the pointer of the first word.

def scramble(s1,s2)
  a1 = s1.split("").sort
  a2 = s2.split("").sort
  pointer1 = 0
  pointer2 = 0

  while pointer2 < a2.length
    return false if pointer1 >= a1.length

    if a1[pointer1] == a2[pointer2]
      pointer1 += 1
      pointer2 += 1
    elsif a1[pointer1] < a2[pointer2]
      pointer1 += 1
    else
      return false
    end
  end
    true
end

Solution 5 - all iterator

def scramble(s1, s2)
s2.chars.all? { |c| s1.sub!(c, '') }
end

Test cases

p scramble('rkqodlw','world') == true
p scramble('cedewaraaossoqqyt','codewars') == true
p scramble('katas','steak') == false
p scramble('scriptjava','javascript') == true
p scramble('scriptingjava','javascript') == true