How to find word with the greatest number of repeated letters - ruby

Question

My goal is to find the word with greatest number of repeated letters in a given string. For example, "aabcc ddeeteefef iijjfff" would return "ddeeteefef" because "e" is repeated five times in this word and that is more than all other repeating characters.
So far this is what I got, but it has many problems and is not complete:
def LetterCountI(str)
s = str.split(" ")
i = 0
result = []
t = s[i].scan(/((.)\2+)/).map(&:max)
u = t.max { |a, b| a.length <=> b.length }
return u.split(//).count
end
The code I have only finds consecutive patterns; if the pattern is interrupted (such as with "aabaaa", it counts a three times instead of five).

Answer 1

str.scan(/\w+/).max_by{ |w| w.chars.group_by(&:to_s).values.map(&:size).max }
scan(/\w+/) — create an array of all sequences of 'word' characters
max_by{ … } — find the word that gives the largest value inside this block
chars — split the string into characters
group_by(&:to_s) — create a hash mapping each character to an array of all the occurrences
values — just get all the arrays of the occurrences
map(&:size) — convert each array to the number of characters in that array
max — find the largest characters and use this as the result for max_by to examine
Edit: Written less compactly:
str.scan(/\w+/).max_by do |word|
word.chars
.group_by{ |char| char }
.map{ |char,array| array.size }
.max
end
Written less functionally and with less Ruby-isms (to make it look more like "other" languages):
words_by_most_repeated = []
str.split(" ").each do |word|
count_by_char = {} # hash mapping character to count of occurrences
word.chars.each do |char|
count_by_char[ char ] = 0 unless count_by_char[ char ]
count_by_char[ char ] += 1
end
maximum_count = 0
count_by_char.each do |char,count|
if count > maximum_count then
maximum_count = count
end
end
words_by_most_repeated[ maximum_count ] = word
end
most_repeated = words_by_most_repeated.last

Answer 2

I'd do as below :
s = "aabcc ddeeteefef iijjfff"
# intermediate calculation that's happening in the final code
s.split(" ").map { |w| w.chars.max_by { |e| w.count(e) } }
# => ["a", "e", "f"] # getting the max count character from each word
s.split(" ").map { |w| w.count(w.chars.max_by { |e| w.count(e) }) }
# => [2, 5, 3] # getting the max count character's count from each word
# final code
s.split(" ").max_by { |w| w.count(w.chars.max_by { |e| w.count(e) }) }
# => "ddeeteefef"
update
each_with_object gives better result than group_by method.
require 'benchmark'
s = "aabcc ddeeteefef iijjfff"
def phrogz(s)
s.scan(/\w+/).max_by{ |word| word.chars.group_by(&:to_s).values.map(&:size).max }
end
def arup_v1(s)
max_string = s.split.max_by do |w|
h = w.chars.each_with_object(Hash.new(0)) do |e,hsh|
hsh[e] += 1
end
h.values.max
end
end
def arup_v2(s)
s.split.max_by { |w| w.count(w.chars.max_by { |e| w.count(e) }) }
end
n = 100_000
Benchmark.bm do |x|
x.report("Phrogz:") { n.times {|i| phrogz s } }
x.report("arup_v2:"){ n.times {|i| arup_v2 s } }
x.report("arup_v1:"){ n.times {|i| arup_v1 s } }
end
output
user system total real
Phrogz: 1.981000 0.000000 1.981000 ( 1.979198)
arup_v2: 0.874000 0.000000 0.874000 ( 0.878088)
arup_v1: 1.684000 0.000000 1.684000 ( 1.685168)

Answer 3

Similar to sawa's answer:
"aabcc ddeeteefef iijjfff".split.max_by{|w| w.length - w.chars.uniq.length}
=> "ddeeteefef"
In Ruby 2.x, this works as-is because String#chars returns an array. In earlier versions of ruby, String#chars yields an enumerator so you need to add .to_a before applying uniq. I did my testing in Ruby 2.0, and overlooked this until it was pointed out by Stephens.
I believe this is valid, since the question was "greatest number of repeated letters in a given string" rather than greatest number of repeats for a single letter in a given string.

Answer 4

"aabcc ddeeteefef iijjfff"
.split.max_by{|w| w.chars.sort.chunk{|e| e}.map{|e| e.last.length}.max}
# => "ddeeteefef"