fail?

solution1('123')
=> 6
solution2('123')
=> 6
solution3('123')
=> 150

EmBRACE the curly braces...

awesome!

amazing👍

Really...?...here you go ;-)

Clever, I'm not sure how it gets any faster than that.

Code here if you want to try:

require 'benchmark'

strings = []

10000.times do 
  string = Array.new(rand(500..1000)) { rand(10).to_s }.join('')
  strings << string
end

def solution1(array)
  array.each do |s|
    s.to_i.digits.sum
  end
end

def solution2(array)
  array.each do |s|
    s.chars.sum(&:to_i)
  end
end

puts Benchmark.measure{ solution1(strings) }
puts Benchmark.measure{ solution2(strings) }

His solution is actually much faster than mine for larger data sets and longer numbers. Benchmark testing on Repl.it given the following parameters:

10,000 strings of length L where 500 <= L < 1,000

string.to_i.digits.sum -- 21.563347 seconds

string.chars.sum(&:to_i) -- 3.311112 seconds

However, my solution is like 2 or 3 characters shorter so clearly mine is more efficient!

Using Repel.it I tested both solutions on over 100
iterations of the test cases and the results are:

string.to_i.digits.sum: 1.1976970482

string.chars.sum(&:to_i): 1.2447343944

So while your solution is unique, the times for
both solutions are about the same.

If n was type of number, spread operator would not work with it (would throw error). Spread operators only work with arrays and string.

BTW, your solution was clever and simple

...for some reason, I thought n is of type number