# Scan array for 3 consecutive numbers that sum to 7

Write a function lucky_sevens?(numbers), which takes in an array of integers and returns true if any three consecutive elements sum to 7.

lucky_sevens?([2,1,5,1,0]) == true # => 1 + 5 + 1 == 7
lucky_sevens?([0,-2,1,8]) == true # => -2 + 1 + 8 == 7
lucky_sevens?([7,7,7,7]) == false
lucky_sevens?([3,4,3,4]) == false


Make sure your code correctly checks for edge cases (i.e. the first and last elements of the array).

While my code does seem to work (I did some test cases), I know it can be improved. There's no reason to have double puts and return statements. I'm also wondering if there are general improvements to the syntax itself that can make it much more readable.

def lucky_sevens?(numbers)
i = 0
while i < numbers.length - 2

if numbers[i] + numbers[i + 1] + numbers[i + 2] == 7
puts true
return true
end
i+=1
end
puts false
return false
end


Your code, is way too imperative to be idiomatic Ruby.

The only reasonable improvement is re-writing it from scratch:

def lucky_sevens?(numbers)
groups_of_three(numbers)
.any? {|group| group.inject(:+) == 7}
end


Knowing that inject(:+) means sum, this reads like English. Implementing groups_of_three is left as an exercise for the reader.

• Could the downvoter explain why? I am always happy to receive constructive criticism :) Oct 26, 2015 at 14:16

Looping using an index is not idiomatic in Ruby. Looping using #each and Enumerable methods based on #each are much more common. This is a perfect opportunity to use Enumerable#each_cons(3).

def lucky_sevens?(numbers)
numbers.each_cons(3) { |group| return true if group.reduce(:+) == 7 }
false
end


The problem doesn't ask you to print anything, so don't.

• Writing groups_of_three was left as an exercise for the reader :o, showing the right built-in right away in my opinion seems to encourage laziness... Oct 20, 2015 at 21:49
• Don't use my return true if …. @Caridorc's any? is much better. Oct 20, 2015 at 21:51
• @200_success, In fairness, your version is more performant as it will return immediately upon finding a success, whereas the any? version requires all groups of three to be calculated, even if, say, the first 3 numbers in the array summed to 7 Oct 21, 2015 at 17:40
• Also, depending on your context, the false return value at the end may be unnecessary, since each_cons returns nil, which will evaluate to false. Oct 21, 2015 at 17:44
• @Jonah Enumeration#any? does short-circuit. You can see that [3,2,1,0].each_cons(3).any? { |group| 1 / group.min } succeeds, while [2,1,0].each_cons(3).any? { |group| 1 / group.min } crashes. Oct 21, 2015 at 17:46