Related
Problem:
Given an array of numbers in Ruby, return the groups of numbers that appear between 1 and 2.
The numbers 1 and 2 do not appear in between other 1's and 2's (there are no subsets of subsets).
Example 1
input: [1, 3, 2, 1, 4, 2]
output: [[1, 3, 2], [1, 4, 2]]
Example 2
input: [0, 1, 3, 2, 10, 1, 5, 6, 7, 8, 7, 5, 2, 3, 1, -400, 2, 12, 16]
output: [ [1, 3, 2], [1, 5, 6, 7, 8, 7, 5, 2], [1, -400, 2] ]
My hunch is to use a combination of #chunk and #drop_while or a generator.
Thanks in advance.
This is an option using [Enumerable#slice_when][1]:
ary1 = [1, 3, 2, 1, 4, 2]
ary2 = [0, 1, 3, 2, 10, 1, 5, 6, 7, 8, 7, 5, 2, 3, 1, -400, 2, 12, 16]
For example:
stop = [1, 2]
ary2.slice_when{ |e| stop.include? e }
.each_slice(2).map { |a, b| b.unshift(a.last) if b }
.reject { |e| e.nil? || (e.intersection stop).empty? }
#=> [[1, 3, 2], [1, 5, 6, 7, 8, 7, 5, 2], [1, -400, 2]]
Other option
More verbose but clearer, given the input:
input = %w(b a b c a b c a c b c a c a)
start = 'a'
stop = 'b'
Using Enumerable#each_with_object, why not use the good old if then else?:
tmp = []
pickup = false
input.each_with_object([]) do |e, res|
if e == start
pickup = true
tmp << e
elsif pickup && e == stop
tmp << e
res << tmp
tmp = []
pickup = false
elsif pickup
tmp << e
end
end
#=> [["a", "b"], ["a", "b"], ["a", "c", "b"]]
[1]: https://ruby-doc.org/core-2.7.0/Enumerable.html#method-i-slice_when
Sounds like an interview question. I'll explain the simplest algorithm I can think of:
You loop through the array once and build the output as you go. When you encounter 1, you store it and the subsequent numbers into another temporary array. When you encounter 2, you put the array in the output array. The edge cases are:
another 1 after you start building the temporary array
a 2 when you don't have a temporary array
First case is easy, always build a new temp array when you encounter a 1. For the second one, you have to check whether you have any items in your temporary array and only append the temp array to your output if it's not empty.
That should get you started.
You could use chunk and Ruby's flip-flop operator:
input = [0, 1, 3, 2, 10, 1, 5, 6, 7, 8, 7, 5, 2, 3, 1, -400, 2, 12, 16]
input.chunk { |i| true if i==1..i==2 }.each { |_, ary| p ary }
Output:
[1, 3, 2]
[1, 5, 6, 7, 8, 7, 5, 2]
[1, -400, 2]
For all people wanting to take a walk on the beach but for obvious reasons can't:
class Flipflop
def initialize(flip, flop) #flip and flop being boolean-returning lambdas
#state = false
#flip = flip
#flop = flop
end
def flipflop(x) #logic taken from The Ruby Programming Language page 111
if !#state
result = #flip[x]
if result
#state = !#flop[x]
end
result
else
#state = !#flop[x]
true
end
end
end
ff = Flipflop.new( ->(x){x == 1}, ->(x){x == 2} )
input = [0, 1, 3, 2, 10, 1, 5, 6, 7, 8, 7, 5, 2, 3, 1, -400, 2, 12, 16]
res = input.select{|el| ff.flipflop(el) }.slice_before(1) #an Enumerator
p res.to_a
# =>[[1, 3, 2], [1, 5, 6, 7, 8, 7, 5, 2], [1, -400, 2]]
For strings, ff = Flipflop.new( ->(x){x.chomp == "BEGIN"}, ->(x){x.chomp == "END"} ) or something like that should work.
Since you commented and added that you are actually reading a file, I deleted my old answer (which was faulty anyways, as #Stefan pointed out) and cam up with this. You can paste this in a file and run it, the DATA IO contains everything that appears after __END__. In your application you would replace it with your File.
class Chunker
BEGIN_INDICATOR = "BEGIN"
END_INDICATOR = "END"
def initialize(io)
#io = io
end
def each
return enum_for(:each) if !block_given?
chunk = nil
while !io.eof? do
line = io.readline.chomp
if line == BEGIN_INDICATOR
chunk = []
chunk << line
elsif line == END_INDICATOR
chunk << line
yield chunk.freeze
chunk = nil
elsif chunk
chunk << line
end
end
end
private
attr_reader :io
end
chunker = Chunker.new(DATA)
chunker.each do |chunk|
p chunk
end
# or, thanks to the `return enum_for(:each) if !block_given?` line:
chunker.each.with_index do |chunk, index|
p "at #{index} is #{chunk}"
end
__END__
ignore
BEGIN
some
thing
END
BEGIN
some
other
thing
END
maybe ignore as well
´´´
You could enhance it to throw EOF when `each` is called multiple times or whatever suits your needs.
So I am trying to get a solution to my two sum problem and I am stuck, I need to print the indices for the elements which add up to the target and my solution will return an element twice if it is one half of the target
def two_sum(nums, target)
num_hash = Hash.new(0)
nums.each_with_index do |num,idx|
num_hash[num] = idx
if num_hash.key?(target - num) && target % num != 0
return [num_hash[num], idx]
end
end
end
So I don't think the problem is related to the number being 1/2 of the target, it just seems to be "if a solution is found, it returns the same index twice". For instance, using the sample set [2, 7, 11, 15]
two_sum([2, 7, 11, 15], 14) # => [2, 7, 11, 15]
So, 7 is half of 14, which is the target, and instead of returning the index 1 twice, as you suggest it would, it returns the original input array (the result of nums.each_with_index. However, if we try passing a target of 9, it behaves as you describe:
two_sum([2, 7, 11, 15], 9) # => [1, 1]
The reason for this, is because of the line:
return [num_hash[num], idx]
you have already set num into the num_hash (num_hash[num] = idx) and then you are returning both the idx and num_hash[num], which is also idx. So what you want to do is:
return [num_hash[target - num], idx]
and then to 'fix' all the elements being returned when no result is found, just return [] at the end of the method:
def two_sum(nums, target)
num_hash = Hash.new(0)
nums.each_with_index do |num,idx|
num_hash[num] = idx
if num_hash.key?(target - num) && target % num != 0
return [num_hash[target - num], idx]
end
end
[]
end
and now:
two_sum([2, 7, 11, 15], 14) # => []
two_sum([2, 7, 11, 15], 9) # => [0, 1]
Note: you also have a problem with the code where, if you have the same number twice, it doesn't find the answer:
two_sum([2, 7, 11, 7, 15], 14) # => []
left for you to figure out, just wanted to point this out to you.
You can use the method Array#combination to advantage here.
def two_sum(nums, target)
nums.each_index.to_a.combination(2).select { |i,j| nums[i] + nums[j] == target }
end
two_sum([2, 7, 11, 15], 14)
#=> []
two_sum([2, 7, 11, 15], 9)
#=> [[0, 1]]
two_sum([2, 4, 7, 5], 9)
#=> [[0, 2], [1, 3]]
two_sum([2, 2, 2, 2], 4)
#=> [[0, 1], [0, 2], [0, 3], [1, 2], [1, 3], [2, 3]]
two_sum([2, 4, 7, 5], 8)
#=> []
For
nums = [2, 4, 7, 5]
target = 9
the steps are as follows.
a = nums.each_index
#=> #<Enumerator: [2, 4, 7, 5]:each_index>
We can see the elements that will be generated by this enumerator by converting it to an array.
b = a.to_a
#=> [0, 1, 2, 3]
Next,
c = b.combination(2)
#=> #<Enumerator: [0, 1, 2, 3]:combination(2)>
c.to_a
#=> [[0, 1], [0, 2], [0, 3], [1, 2], [1, 3], [2, 3]]
The rest is straightforward as select merely selects those pairs of indices passed to it (i,j) whose corresponding values, num[i] and num[j], sum to target.
I think what you want is ...
return [num_hash[target-num], idx]
This is what I am trying to get:
Array1=[a,b,c]
Array2=[d,e,f]
=> [a*d,a*e,a*f,b*d,b*e,b*f,c*d,c*e,c*f]
How can you do this in ruby? So far I can only return [a*d,b*e,c*f]
Read Array#product method.
array1 = [1, 2, 3]
array2 = [3, 5, 6]
array1.product(array2).map { |a, b| a * b }
# => [3, 5, 6, 6, 10, 12, 9, 15, 18]
Array1.product(Array2).map{|x, y| x * y}
I need to add consecutive numbers to a new array and, if it is not a consecutive number, add only that value to a new array:
old_array = [1, 2, 3, 5, 7, 8, 9, 20, 21, 23, 29]
I want to get this result:
new_array = [
[1,2,3],
[5],
[7,8,9]
[20,21]
[23],
[29]
]
Is there an easier way to do this？
A little late to this party but:
old_array.slice_when { |prev, curr| curr != prev.next }.to_a
# => [[1, 2, 3], [5], [7, 8, 9], [20, 21], [23], [29]]
This is the official answer given in RDoc (slightly modified):
actual = old_array.first
old_array.slice_before do
|e|
expected, actual = actual.next, e
expected != actual
end.to_a
A couple other ways:
old_array = [1, 2, 3, 5, 7, 8, 9, 20, 21, 23, 29]
#1
a, b = [], []
enum = old_array.each
loop do
b << enum.next
unless enum.peek.eql?(b.last.succ)
a << b
b = []
end
end
a << b if b.any?
a #=> [[1, 2, 3], [5], [7, 8, 9], [20, 21], [23], [29]]
#2
def pull_range(arr)
b = arr.take_while.with_index { |e,i| e-i == arr.first }
[b, arr[b.size..-1]]
end
b, l = [], a
while l.any?
f, l = pull_range(l)
b << f
end
b #=> [[1, 2, 3], [5], [7, 8, 9], [20, 21], [23], [29]]
Using chunk you could do:
old_array.chunk([old_array[0],old_array[0]]) do |item, block_data|
if item > block_data[1]+1
block_data[0] = item
end
block_data[1] = item
block_data[0]
end.map { |_, i| i }
# => [[1, 2, 3], [5], [7, 8, 9], [20, 21], [23], [29]]
Some answers seem unnecessarily long, it is possible to do this in a very compact way:
arr = [1, 2, 3, 5, 7, 8, 9, 20, 21, 23, 29]
arr.inject([]) { |a,e| (a[-1] && e == a[-1][-1] + 1) ? a[-1] << e : a << [e]; a }
# [[1, 2, 3], [5], [7, 8, 9], [20, 21], [23], [29]]
Alternatively, starting with the first element to get rid of the a[-1] condition (needed for the case when a[-1] would be nil because a is empty):
arr[1..-1].inject([[arr[0]]]) { |a,e| e == a[-1][-1] + 1 ? a[-1] << e : a << [e]; a }
# [[1, 2, 3], [5], [7, 8, 9], [20, 21], [23], [29]]
Enumerable#inject iterates all elements of the enumerable, building up a result value which starts with the given object. I give it an empty Array or an Array with the first value wrapped in an Array respectively in my solutions. Then I simply check if the next element of the input Array we are iterating is equal to the last value of the last Array in the resulting Array plus 1 (i.e, if it is the next consecutive element). If it is, I append it to the last list. Otherwise, I start a new list with that element in it and append it to the resulting Array.
You could also do it like this:
old_array=[1, 2, 3, 5, 7, 8, 9, 20, 21, 23, 29]
new_array=[]
tmp=[]
prev=nil
for i in old_array.each
if i != old_array[0]
if i - prev == 1
tmp << i
else
new_array << tmp
tmp=[i]
end
if i == old_array[-1]
new_array << tmp
break
end
prev=i
else
prev=i
tmp << i
end
end
Using a Hash you can do:
counter = 0
groups = {}
old_array.each_with_index do |e, i|
groups[counter] ||= []
groups[counter].push old_array[i]
counter += 1 unless old_array.include? e.next
end
new_array = groups.keys.map { |i| groups[i] }
How can I substitue an element in an array?
a = [1,2,3,4,5]
I need to replace 5 with [11,22,33,44].flatten!
so that a now becomes
a = [1,2,3,4,11,22,33,44]
Not sure if you're looking to substitute a particular value or not, but this works:
a = [1, 2, 3, 4, 5]
b = [11, 22, 33, 44]
a.map! { |x| x == 5 ? b : x }.flatten!
This iterates over the values of a, and when it finds a value of 5, it replaces that value with array b, then flattens the arrays into one array.
Perhaps you mean:
a[4] = [11,22,33,44] # or a[-1] = ...
a.flatten!
A functional solution might be nicer, how about just:
a[0..-2] + [11, 22, 33, 44]
which yields...
=> [1, 2, 3, 4, 11, 22, 33, 44]
The version of bta using a.index(5) is the fastest one:
a[a.index(5)] = b if a.index(5) # 5.133327 sec/10^6
At least 10% faster than Ryan McGeary's one:
a.map!{ |x| x == 5 ? b : x } # 5.647182 sec/10^6
However, note that a.index(5) only return the first index where 5 is found.
So, given an array where 5 appears more than once, results will be different:
a = [1, 2, 3, 4, 5, 5]
b = [11,22,33,44]
a[a.index(5)] = b if a.index(5)
a.flatten # => [1, 2, 3, 4, 11, 22, 33, 44, 5]
a.map!{ |x| x == 5 ? b : x }
a.flatten # => [1, 2, 3, 4, 11, 22, 33, 44, 11, 22, 33, 44]
Array#delete will return the item or nil. You may use this to know whether or not to push your new values
a.push 11,22,33,44 if a.delete 5
You really don't have to flatten if you just concatenate. So trim the last element off the first array and concatenate them:
a = [ 1, 2, 3, 4, 5 ] #=> [1, 2, 3, 4, 5]
t = [11, 22, 33, 44] #=> [11, 22, 33, 44]
result = a[0..-2] + t #=> [1, 2, 3, 4, 11, 22, 33, 44]
a[0..-2] is a slice operation that takes all but the last element of the array.
Hope it helps!
This variant will find the 5 no matter where in the array it is.
a = [1, 2, 3, 4, 5]
a[a.index(5)]=[11, 22, 33, 44] if a.index(5)
a.flatten!
Here is another simple way to replace the value 5 in the array:
a[-1, 1] = [11, 22, 33, 44]
This uses the Array#[]= method. I'm not exactly sure why it works though.
gweg, not sure what you're trying to do here, but are you looking for something like this?
a = [1, 2, 3, 4, 5]
a.delete_at(4)
a = a.concat([11,22,33,44])
There are a number of ways of doing this -- I don't think the code above is especially nice looking. It all depends on the significance of '5' in your original array.