How to remove 3 or more the same character - ruby

Question

During my lessons of Ruby I came across of this exercise. I'm trying to remove 3 or more the same charactes in a row. Test Cases Input: abbbaaccada Output: ccada Input: bbccdddcb Output: (Empty string)
So far I have solution which doesn't return expected results:
def playground("abbbaaccada")
count = string.length
string.chars.each_with_index.map { |v, i| (v * (count - i)).capitalize }.join('')
end
output gives me
==> AaaaaaaaaaaBbbbbbbbbbBbbbbbbbbBbbbbbbbAaaaaaaAaaaaaCccccCcccAaaDdA
instead of
==> ccada
Could you please advise?
Edit:
Forgot to add that regexp isn't allowed

Answer 1

There are two challenges here:
Match and remove any run of thee or more characters in a row
Recurse to test again in case the previous step created a new run of three
Here's one way to do it:
THREE_OR_MORE = /(.)\1{2,}/
def three_is_too_many(str)
if str.match? THREE_OR_MORE
str = three_is_too_many(str.gsub(THREE_OR_MORE, ''))
end
str
end
The regexp finds any character ('.'), followed by itself ('\1'), two or more times ('{2,}').
Then the routine either a) removes three or more and tests again or b) returns the string.

Answer 2

Here's a potential solution. The following method searches for any subsequence of an array with repeats, and returns the range of the repeated values if there are three or more of them.
def find_3_or_more(ary)
ary.each_index do |i|
j = i + 1
while j < ary.length && ary[i] == ary[j]
j += 1
end
return (i...j) if j - i > 2
end
nil
end
This portion breaks the target string into an array of chars, and repeatedly slices out the characters in ranges identified as repeats until there are none, as indicated by a nil range.
def delete_3_or_more(str)
ary = str.chars
while r = find_3_or_more(ary)
ary.slice!(r)
end
return ary.join
end
It seems to do the job for your test cases.

Answer 3

def recursively_remove_runs_of_3_or_more(str)
arr = str.chars
loop do
a = arr.slice_when { |a,b| a.downcase != b.downcase }.to_a
b = a.reject! { |e| e.size > 2 }
arr = a.flatten
break arr.join if b.nil?
end
end
recursively_remove_runs_of_3_or_more "abbbaaccada"
#=> "ccada"
This uses Enumerable#slice_when (new in MRI v2.2). Note that Array#reject! returns nil when no changes are made.
You could alternatively use Enumerable#chunk_while (new in MRI v2.3). Simply replace:
a = arr.slice_when { |a,b| a.downcase != b.downcase }.to_a
with:
a = arr.chunk_while { |a,b| a.downcase == b.downcase }.to_a
chunk_while and slice_when are yin and yang.
If a regular expression could be used and case where not an issue, you could write:
str = "abbbaaccada"
s = str.dup
loop { break(s) if s.gsub!(/(.)\1{2,}/, '').nil? }
#=> "ccada"

Answer 4

(I wanted to comment, but it doesn't allow me to do that yet.)
Assuming that you are trying to learn, I chose to only give you some tips while avoiding a solution.
There might be shorter ways of doing this using regex or/and some String methods. However, you said you can not use regex.
My tip is, try to solve it only using the sections you have covered so far. It may not necessarily be the most elegant solution, but you can revise it as you progress. As others suggested, recursion might be a good option. But, if you are not familiar with that yet, you can try slicing the string and merging the parts you need. This can be combined with an endless loop to check the new string satisfies your condition: but think about when you need to break out of the loop.
Also, in your code:
v * (count - i)
String#* actually gives you count - i copies of v, concatenated together.