Get max consecutive occurrences of value in array

Question

Is there a more elegant way to achieve this below:

Input:

array = [1, 1, 1, 0, 0, 1, 1, 1, 1, 0]

Output:

4

My algo:

streak = 0
max_streak = 0

arr.each do |n|
  if n == 1
    streak += 1
  else
    max_streak = streak if streak > max_streak
    streak = 0
  end
end

puts max_streak

Answer 1

Similar to w0lf's answer, but skipping elements by returning nil from chunk:

array.chunk { |x| x == 1 || nil }.map { |_, x| x.size }.max

Answer 2

Edit: Another way to do this (that is less generic than Stefan's answer since you would have to flatten and split again if there was another number other than 0 and 1 in there, but easier to use in this case):

array.split(0).max.count

---

You can use:

array.chunk { |n| n }.select { |a| a.include?(1) }.map { |y, ys| ys.count}.max

ref: Count sequential occurrences of element in ruby array

Answer 3

You can use Enumerable#chunk:

p array.chunk{|x| x}.select{|x, xs| x == 1}.map{|x, xs| xs.size }.max

This is more concise, but if performance was important, I'd use your approach.

---

Edit: If you're in Ruby 2.2.2, you can also use the new Enumerable#slice_when method (assuming your input array consists of only 0s and 1s):

array.slice_when{|x,y| x < y }.map{|slice| slice.count 1 }.max