I need to have true in all 3 tests. I need to iterate the code until the array returns only the longest ones only.
# Tests
p longest(['tres', 'pez', 'alerta', 'cuatro', 'tesla', 'tropas', 'siete']) == ["alerta", "cuatro", "tropas"]
p longest(['gato', 'perro', 'elefante', 'jirafa']) == ["elefante"]
p longest(['verde', 'rojo', 'negro', 'morado']) == ["morado"]
Okay, let's break it down...
First: find out how long the longest word is:
words = ['tres', 'pez', 'alerta', 'cuatro', 'tesla', 'tropas', 'siete']
words.map(&:length).max
#=> 6
Second select all words with that length:
words.select { |w| w.length == 6 }
#=> ["alerta", "cuatro", "tropas"]
Combine that to a method:
def longest(words)
max_length = words.map(&:length).max
words.select { |w| w.length == max_length }
end
Related
This function should take in two strings "daBcD" and "ABC". It is trying to create the string "b" from the letters in "a". You can only delete or capitalize letters, you cant change them. b will always contain all uppercase letters.
def abbreviation(a, b)
aArray = a.split('')
idx = 0
aArray.each do |char|
#print "char: #{char}\n"
#print "Before loops: #{aArray}\n"
if char.casecmp(b[idx]) == 0
char.upcase!
idx += 1
#print "char: #{char}\nArry: #{aArray}\n"
#print "idx: #{idx}\n siz: #{b.size}\n"
if idx == b.size
aArray.reject! {|i| i == 'delete'}
aArray.slice!(b.size)
break
end
else
aArray[aArray.index(char)] = 'delete'
#print "deleted, now is: #{aArray}\n"
end
end
res = aArray.join('')
if res == b
return 'YES'
else
return 'NO'
end
end
This works for a couple test cases, but fails most of them. Can someone describe a better approach?
I have assumed the problem is to determine whether the characters in b appear in a (case indifferent), in the same order as in b, but not necessarily contiguous in a (see the second example below). If they do I return an array of the indices at which they appear in a. If there is no match, nil is returned.
def doit(a, b)
m = a.match(Regexp.new(b.each_char.map { |c| "(#{c})" }.join('.*'),
Regexp::IGNORECASE))
return nil if m.nil?
(1..b.size).map { |i| m.begin(i) }
end
doit "daBcD", "ABC"
#=> [1, 2, 3]
doit "daXBDecf", "ABC"
#=> [1, 3, 6]
doit "dacBD", "ABC"
#=> nil
For the first example the regular expression is as follows.
Regexp.new("ABC".each_char.map { |c| "(#{c})" }.join('.*'), Regexp::IGNORECASE)
#=> /(A).*(B).*(C)/i
The absolutely easiest way is via regular expression:
def abbreviation(a, b)
re = Regexp.new(b.each_char.map(&Regexp.method(:quote)).join('.*'), Regexp::IGNORECASE)
!!re.match(a)
end
abbreviation("daBcD", "ABC")
# => true
abbreviation("daCbD", "ABC")
# => false
For the input ABC, we'll construct a regular expression /A.*B.*C/i, then test the other string against it. The .* construct will account for "deletion"; the IGNORECASE option for "capitalisation".
EDIT: If the problem is further constrained that only lowercase letters can be deleted, as suggested by the comments,
def abbreviation(a, b)
# (b is uppercase only)
re_pat = b.each_char.map { |c| "[#{c}#{c.downcase}]"}.join('[[:lower:]]*')
re = Regexp.new(re_pat)
!!re.match(a)
end
p abbreviation("daBcD", "ABC") # => true
p abbreviation("daBdcD", "ABC") # => true
p abbreviation("daBDcD", "ABC") # => false
Here is my code in ruby for a word compression.
For any given word (e.g. abbbcca) the compressed word/output should be in the format as "letter+repetition" (for above example, output: a1b3c2a1).
Here I'm so close to the completion but my result isn't in the expected format. It's counting the whole letters in string.chars.each thus resulting output as a2b3c2a2.
Any help?
def string_compressor(string)
new_string = []
puts string.squeeze
string.squeeze.chars.each { |s|
count = 0
string.chars.each { |w|
if [s] == [w]
count += 1
end
}
new_string << "#{s}#{count}"
puts "#{new_string}"
}
if new_string.length > string.length
return string
elsif new_string.length < string.length
return new_string
else "Equal"
end
end
string_compressor("abbbcca")
'abbbcca'.chars.chunk{|c| c}.map{|c, a| [c, a.size]}.flatten.join
Adapted from a similar question.
Similar:
'abbbcca'.chars.chunk{|c| c}.map{|c, a| "#{c}#{a.size}"}.join
See chunk documentation
You can use a regular expression for that.
'abbbcca'.gsub(/(.)\1*/) { |m| "%s%d" % [m[0], m.size] }
#=> "a1b3c2a1"
The regular expression reads, "match any character, capturing it in group 1. Then match the contents of capture group 1 zero or more times".
As you said, your code counts every letter in the string, not just the one grouped next to one another.
Here's a modified version :
def display_count(count)
if count == 1
""
else
count.to_s
end
end
def string_compressor(string)
new_string = ''
last_char = nil
count = 0
string.chars.each do |char|
if char == last_char
count += 1
else
new_string << "#{last_char}#{display_count(count)}" if last_char
last_char = char
count = 1
end
end
new_string << "#{last_char}#{display_count(count)}" if last_char
new_string
end
p string_compressor('abbbcca') #=> "ab3c2a"
p string_compressor('aaaabbb') #=> "a4b3"
p string_compressor('aabb') #=> "a2b2"
p string_compressor('abc') #=> "abc"
Note that with display_count removing 1s from the string, new_string can never be longer than string. It also probably isn't a good idea to return Equal as a supposedly compressed string.
To decompress the string :
def string_decompressor(string)
string.gsub(/([a-z])(\d+)/i){$1*$2.to_i}
end
p string_decompressor("a5b11") #=> "aaaaabbbbbbbbbbb"
p string_decompressor("ab3c2a") #=> "abbbcca"
I want to make a word unscarambler in ruby. say if I have a words in array
words = ["foo","ofo"]
how can I compare this to another string like "oof" and returning true value
This can be done as follows.
words = ["foo", "ofo", "goo"]
target = "foo"
target_size = target.size
#=> 3
target_sorted = target.each_char.sort
#=> ["f", "o", "o"]
words.select { |w| anagram?(target_size, target_sorted, w) }
#=> ["foo", "ofo"]
The typical way anagram? is written is:
def anagram?(target_size, target_sorted, w)
return false unless w.size == target_size
w.each_char.sort == target_sorted
end
However, I've wondered it might be faster to:
Step through the characters of target
Search for the index i of a matching character in w
If a match is found, delete w[i]
if no match is found (i #=> nil), return false
return true if false is not returned earlier
This can be implemented thus:
def anagram?(target_size, target, w)
return false unless target.size == w.size
wcpy = w.dup
target.each_char do |c|
i = wcpy.index(c)
return false unless i
wcpy[i] = ''
end
true
end
words.select { |w| anagram?(target_size, target, w) }
#=> ["foo", "ofo"]
I'll have to benchmark the two one day.
We could also write:
def anagram?(w1, w2)
return false unless w1.size == w2.size
w1.chars.difference(w2.chars).empty?
end
The helper Array#difference is defined here.
If all the strings in the array are permutations of each other then:
words = ["foo", "ofo"]
str = "foo"
words[0].split("").sort == str.split("").sort
I've seen the solution and it more or less matches
Write a method that takes a string and returns the number of vowels
in the string. You may assume that all the letters are lower cased. You can treat "y" as a consonant.
Difficulty: easy.
def count_vowels(string)
vowel = 0
i = 0
while i < string.length
if (string[i]=="a" || string[i]=="e" || string[i]=="i" || string[i]=="o"|| string[i]=="u")
vowel +=1
end
i +=1
return vowel
end
puts("count_vowels(\"abcd\") == 1: #{count_vowels("abcd") == 1}")
puts("count_vowels(\"color\") == 2: #{count_vowels("color") == 2}")
puts("count_vowels(\"colour\") == 3: #{count_vowels("colour") == 3}")
puts("count_vowels(\"cecilia\") == 4: #{count_vowels("cecilia") == 4}")
def count_vowels(str)
str.scan(/[aeoui]/).count
end
/[aeoui]/ is a regular expression that basically means "Any of these characters: a, e, o, u, i". The String#scan method returns all matches of a regular expression in the string.
def count_vowels(str)
str.count("aeoui")
end
Your function is fine you are just missing a keyword end to close of your while loop
def count_vowels(string)
vowel = 0
i = 0
while i < string.length
if (string[i]=="a" || string[i]=="e" || string[i]=="i" || string[i]=="o"|| string[i]=="u")
vowel +=1
end
i +=1
end
return vowel
end
puts("count_vowels(\"abcd\") == 1: #{count_vowels("abcd") == 1}")
puts("count_vowels(\"color\") == 2: #{count_vowels("color") == 2}")
puts("count_vowels(\"colour\") == 3: #{count_vowels("colour") == 3}")
puts("count_vowels(\"cecilia\") == 4: #{count_vowels("cecilia") == 4}")
#=> count_vowels("abcd") == 1: true
#=> count_vowels("color") == 2: true
#=> count_vowels("colour") == 3: true
#=> count_vowels("cecilia") == 4: true
I think using HashTable data structure would be good way to go for this particular problem. Especially if you're required to output number of every single vowel separately.
Here is the code I'd use:
def vowels(string)
found_vowels = Hash.new(0)
string.split("").each do |char|
case char.downcase
when 'a'
found_vowels['a']+=1
when 'e'
found_vowels['e']+=1
when 'i'
found_vowels['i']+=1
when 'o'
found_vowels['o']+=1
when 'u'
found_vowels['u']+=1
end
end
found_vowels
end
p vowels("aeiou")
Or even this (elegant but not necessarily performant):
def elegant_vowels(string)
found_vowels = Hash.new(0)
string.split("").each do |char|
case char.downcase
when ->(n) { ['a','e','i','o','u'].include?(n) }
found_vowels[char]+=1
end
end
found_vowels
end
p elegant_vowels("aeiou")
which would output:
{"a"=>1, "e"=>1, "i"=>1, "o"=>1, "u"=>1}
So you don't have to turn the string into an array and worry about case sensitivity:
def getVowelCount(string)
string.downcase.count 'aeiou'
end
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).
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
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)
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.
"aabcc ddeeteefef iijjfff"
.split.max_by{|w| w.chars.sort.chunk{|e| e}.map{|e| e.last.length}.max}
# => "ddeeteefef"